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Limit Properties

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The time has almost come for us to actually compute some limits. However, before we do that we will need some properties of limits that will make our life somewhat easier. So, let’s take a look at those first. The formal proofs of these properties depend on the mathematical definition of the limit, which we will save for the end of the chapter. For now, we will offer some intuition rather than proof for many of the properties.

Limit Properties

First, we will assume that $\lim_ \limits{x\to a} f(x)$ and $\lim_\limits{x \to a} g(x)$ exist and that $c$ is any constant. Then,

  1. $\lim_\limits{x\to a} [c~f(x)] = c\lim_\limits{x\to a} f(x)$.
    2. In other words, we can "factor" a multiplicative constant out of a limit.

    Suppose that $\lim_\limits{x\to a} f(x) = L$. Using our informal language, this means that as our x values get very close to $a$, the values of $f(x)$ (that is, the $y$ values, or height of the function) get closer and closer to $L$. In fact, we can make the height as close to $L$ as we want as long as we pick a value of $x$ that is close enough to $a$.

    Now, let's think about $\lim_\limits{x\to a} c\cdot f(x)$. That $c$ value will scale the graph of $f(x)$ vertically by a factor of $c$. That means that now, as we let our x values get closer to $a$, all of the heights will be multiplied by a factor of $c$. So, as $x$ gets closer to $a$, $c\cdot f(x)$ will get very close to $c$ times the original limit. Therefore $\lim_\limits{x\to a} cf(x) = c\cdot L = c\cdot \lim_\limits{x\to a} f(x)$.



  2. $\lim_\limits{x\to a}[f(x)\pm g(x)] = \lim_\limits{x\to a} f(x) \pm \lim_\limits{x\to a} g(x)$
    3. So, to take the limit of a sum or difference all we need to do is take the limit of the individual parts and then put them back together with the appropriate sign. This is also not limited to two functions. This fact will work no matter how many functions we’ve got separated by “+” or “-”.

  3. $\lim_\limits{x\to a}[f(x)g(x)] = \lim_\limits{x\to a}f(x) \lim_\limits{x\to a} g(x)$
    4. We take the limits of products in the same way that we can take the limit of sums or differences. Just take the limit of the pieces and then put them back together. Also, as with sums or differences, this fact is not limited to just two functions.

  4. $\lim_\limits{x\to a}\left[\frac{f(x)}{g(x)}\right] = \frac{\lim_\limits{x\to a}f(x)}{\lim_\limits{x\to a} g(x)}$, provided $\lim_\limits{x\to a}g(x)\neq 0$.
    5. As noted in the statement we only need to worry about the limit in the denominator being zero when we do the limit of a quotient. If it were zero we would end up with a division by zero error and we need to avoid that.


These limits can be understood from our intuition and picturing functions, as in the informal proof of the first property above. If we want to prove them more formally, we'll have to wait until we understand the formal definition of the limit.

There are more limit properties to come, but let's see what kinds of things we can do with these properties so far.

Given $\lim_\limits{x\to 8}f(x) = -9, \lim_\limits{x\to 8}g(x)=2$ and $\lim_\limits{x\to 8} h(x) = 4$ use the limit properties to compute the following limit. If it is not possible to compute, clearly explain why not.

$$\lim\limits_{x\to 8}[2f(x)-12h(x)]$$

We'll work on unpeeling this limit, layer by layer, so we'll kind of apply the order of operations in reverse. Structurally, we notice a big minus sign, so we can use limit property 2 to rewrite the limit as $$\lim\limits_{x\to 8} [2f(x)-12h(x)] = \lim\limits_{x\to 8}[2f(x)]-\lim\limits_{x\to 8}[12h(x)]. $$ Next, we can use property 1 to pull each constant multiple out of the limits: $$\lim\limits_{x\to 8}[2f(x)]-\lim\limits_{x\to 8}[12h(x)] = 2\lim\limits_{x\to 8}[f(x)] - 12\lim\limits_{x\to 8}[h(x)].$$ Finally, we can use the given information that $\lim_\limits{x\to 8}f(x) = -9$ and $\lim_\limits{x\to 8}h(x)=4$ to substitute in these quantities and evaluate. $$2\textcolor{blue}{\lim\limits_{x\to 8}[f(x)]} - 12\textcolor{purple}{\lim\limits_{x\to 8}[h(x)]}$$ $$= 2\textcolor{blue}{(-9)} - 12\textcolor{purple}{(4)}$$ $$= -18 - 48 = -66$$ Therefore, $\lim\limits_{x\to 8}[2f(x)-12h(x)] = -66$.


Given $\lim_\limits{x\to 8}f(x) = -9, \lim_\limits{x\to 8}g(x)=2$ and $\lim_\limits{x\to 8} h(x) = 4$ use the limit properties to compute the following limit. If it is not possible to compute, clearly explain why not.

$$\lim\limits_{x\to 8}\left[\frac{f(x)}{g(x)} + \frac{f(x)}{2g(x)-h(x)}\right]$$

Once again, we'll start with property 2, so we split up the limit into two limits with a "+" in between them: $$\lim\limits_{x\to 8}\left[\frac{f(x)}{g(x)} + \frac{f(x)}{2g(x)-h(x)}\right] =\lim\limits_{x\to 8}\frac{f(x)}{g(x)} + \lim\limits_{x\to 8} \frac{f(x)}{2g(x)-h(x)}.$$ At this point, applying property 4 becomes pretty appealing, so that we can deal with those fractions. Remember, though, property 4 has an important caveat: $$\lim\limits_{x\to a} \frac{f(x)}{g(x)} = \frac{\lim\limits_{x\to a} f(x)}{\lim\limits_{x\to a} g(x)} \text{ if } \lim_{x\to a} g(x) \neq 0$$ That is, if we want to use property 4, we need to ensure that the resulting denominator won't be 0. We're okay in the case of the first fraction, since $\lim\limits_{x \to 8} g(x) = 2$, but for the second fraction, we need to investigate what will happen with $\lim\limits_{x\to 8}[2g(x) - h(x)]$. Let's dig in to just that piece of the picture. We'll once more start with property 2: $$\lim\limits_{x\to 8}[2 g(x) - h(x)] =\lim\limits_{x\to 8}[2 g(x)] -\lim\limits_{x\to 8}h(x).$$ Then, for the first term, we can use property 1 to pull out the 2: $$\lim\limits_{x\to 8}[2 g(x)] -\lim\limits_{x\to 8}h(x) =2\lim\limits_{x\to 8}[g(x] - \lim\limits_{x\to 8}[h(x)].$$ Now we substitute in our known limit values and evaluate: $$2\textcolor{blue}{\lim\limits_{x\to 8}[g(x)]} - \textcolor{purple}{\lim\limits_{x\to 8}[h(x)]}$$ $$=2\cdot \textcolor{blue}{2} - \textcolor{purple}{4}$$ $$=4 - 4 = 0$$ Oh no! The bottom limit results in a 0, so we cannot use property 4 to work on our original limit. Therefore, our final answer is that it is not possible to compute this limit with the given information.


Given $\lim_\limits{x\to 8}f(x) = -9, \lim_\limits{x\to 8}g(x)=2$ and $\lim_\limits{x\to 8} h(x) = 4$ use the limit properties to compute the following limit. If it is not possible to compute, clearly explain why not.

$$\lim_{x\to -8} [f(x)g(x)-h(x)]$$

Wait a second, there's something fishy here. Check out that limit again... $$\lim_{\textcolor{blue}{x\to -8}} [f(x)g(x)-h(x)]$$

The original limit information we're given is all about $x\to 8$, but here we have $x \to -8$. We don't know anything about what's happening to these functions as x approaches -8, so we cannot evaluate this limit!



Let's take a look at some more properties and what we can do with them.

More Limit Properties

First, we will assume that $\lim_ \limits{x\to a} f(x)$ and $\lim_\limits{x \to a} g(x)$ exist and that $c$ is any constant. Then,

  1. $\lim\limits_{x\to a} [f(x)]^n = \left[\lim\limits_{x\to a}f(x)\right]^n$ where $n$ is any real number

    2. Note: if $n\lt 0$, we must assume that $\lim\limits_{x\to a}f(x) \neq 0$.

    In this property $n$ can be any real number (positive, negative, integer, fraction, irrational, zero, etc.).


In the case that $n$ is an integer this rule can be thought of as an extended case of property 3. For example, consider $n = 2$. $$\begin{array}{clc} \lim\limits_{x\to a}[f(x)]^2 &= \lim\limits_{x\to a}[f(x)\cdot f(x)] &\\ &= \lim\limits_{x\to a} f(x) \lim\limits_{x \to a} f(x) &\text{ using property 3}\\ &=\left[\lim\limits_{x\to a}f(x)\right]^2\\ \end{array}$$ The same can be done for any integer $n$.


More Limit Properties

First, we will assume that $\lim_ \limits{x\to a} f(x)$ and $\lim_\limits{x \to a} g(x)$ exist and that $c$ is any constant. Then,

  1. $\lim\limits_{x\to a}\left[\sqrt[n]{f(x)}\right] = \sqrt[n]{\lim\limits_{x\to a} f(x)}$.

    2. Note: if $n$ is even, we must assume that $\lim\limits_{x\to a}f(x) \geq 0$.


This is a special case of the previous property, since we can write $$\begin{array}{clc} \lim\limits_{x\to a}\left[\sqrt[n]{f(x)}\right] &= \lim\limits_{x\to a}[f(x)]^{1/n}&\\ &=\left[\lim\limits_{x\to a} f(x)\right]^{1/n}&\text{by property 5}\\ &= \sqrt[n]{\lim\limits_{x\to a}f(x)}&\\ \end{array}$$


Given $\lim\limits_{x \to 0}f(x) = 6,\lim\limits_{x \to 0}g(x) = -4,$ and $\lim\limits_{x \to 0}h(x) = -1$, use the limit properties to compute the following limit. $$\lim\limits_{x \to 0}[f(x)+h(x)]^3$$

Here is the work for this limit. At each step the property (or properties) used are listed and note that in some cases the properties may have been used more than once in the indicated step.

$$\begin{array}{clcl} \lim\limits_{x \to 0}[f(x)+h(x)]^3 &= \left[\lim\limits_{x \to 0}(f(x)+h(x))\right]^3 &~~& \text{Property 5}\\ &= \left[\lim\limits_{x \to 0}f(x) + \lim\limits_{x \to 0}h(x)\right]^3&&\text{Property 2}\\ &= [6 - 1]^3 &&\text{Given limit values}\\ &=125 &&\\ \end{array}$$

Given $\lim\limits_{x \to 0}f(x) = 6,\lim\limits_{x \to 0}g(x) = -4,$ and $\lim\limits_{x \to 0}h(x) = -1$, use the limit properties to compute the following limit. $$\lim\limits_{x \to 0}\sqrt{\frac{f(x)}{h(x)-g(x)}}$$

Here is the work for this limit. At each step the property (or properties) used are listed and note that in some cases the properties may have been used more than once in the indicated step. $$\begin{array}{clcl} \lim\limits_{x \to 0}\sqrt{\frac{f(x)}{h(x)-g(x)}} &= \sqrt{\lim\limits_{x \to 0}\frac{f(x)}{h(x)-g(x)}} &~~&\text{Property 6}\\ \end{array}$$

Now, we're again tempted to use property 4 to handle the fraction, but we need to do some side work to make sure that the denominator won't end up being 0. $$\begin{array}{clcl} \lim\limits_{x \to 0}(h(x)-g(x)) &= \lim\limits_{x \to 0}h(x) - \lim\limits_{x \to 0}g(x) &~~&\text{Property 2}\\ &= -1 - (-4) &&\text{Given limit values}\\ &= 3 &&**\\ \end{array}$$ Good news! The denominator won't be 0, so we can go back to our original limit and continue with property 4.

$$\begin{array}{clcl} \sqrt{\lim\limits_{x \to 0}\frac{f(x)}{h(x)-g(x)}} &= \sqrt{\frac{\lim\limits_{x \to 0}f(x)}{\lim\limits_{x \to 0}(h(x)-g(x))}} &~~&\text{Property 4}\\ &= \sqrt{\frac{6}{3}} && \text{Given limit value and work from **}\\ &=\sqrt{2} &&\\ \end{array}$$


Given $\lim\limits_{x \to 0}f(x) = 6,\lim\limits_{x \to 0}g(x) = -4,$ and $\lim\limits_{x \to 0}h(x) = -1$, use the limit properties to compute the following limit. $$\lim_{x\to 0}\sqrt[3]{g(x)h(x)}$$

Here is the work for this limit. At each step the property (or properties) used are listed and note that in some cases the properties may have been used more than once in the indicated step. $$\begin{array}{clcl} \lim\limits_{x\to 0}\sqrt[3]{g(x)h(x)} &= \sqrt[3]{\lim\limits_{x\to0} g(x)h(x)} &~~ &\text{Property 6}\\ &= \sqrt[3]{\left[\lim\limits_{x\to0}g(x)\right]\left[\lim\limits_{x\to0}h(x)\right]} && \text{Property 3}\\ &=\sqrt[3]{(-4)(-1)} &&\text{Given limit values}\\ &=\sqrt[3]{4}&&\\ \end{array}$$



We've worked with a lot of abstract functions up until this point. Let's check out some functions that we can visualize more specifically.

Even More Limit Properties

First, we will assume that $\lim_ \limits{x\to a} f(x)$ and $\lim_\limits{x \to a} g(x)$ exist and that $c$ is any constant. Then,

  1. $\lim\limits_{x\to a} c = c$, where $c$ is any real number.

    2. In other words, the limit of a constant is just the constant.

    For this proof, we'll look at a graph. Here is the graph of $y = c$ for some real number c:

    a horizontal line at a height of c. The point x = a is marked on the horizontal axis, with a dotted line up to the horizontal line.

    When we find the limit as $x \to a$, we're asking: what happens to the height of the function as $x$ gets very close to $a$? Well, the height of this function is always $c$, so the limit must be $c$!



  2. $\lim\limits_{x\to a} x = a$

    3. Again, let's check out the graph. Here is the graph of the line $y = x$:

    A diagonal line through the origin and the point (a,a). The line is labeled y = x.

    This line goes through points like $(-1,-1)$, $(2,2)$, and $(10,10)$, or more generally, $(a,a)$. So, as our x-values get closer and closer to $a$, the height also gets very close to $a$, meaning $\lim\limits_{x\to a} x = a$.



  3. $\lim\limits_{x\to a}x^n = a^n$

    4. Note: if $n$ is negative, we must assume that $a \neq 0$.

    This is really just a special case of property 5 using $f(x) = x$.

    $$\begin{array}{clcl} \lim\limits_{x\to a}x^n &= \left[\lim\limits_{x\to a} x\right]^n &~~& \text{Property 5}\\ &=(a)^n && \text{Property 8}\\ \end{array}$$

Note that all these properties also hold for the two one-sided limits as well we just didn’t write them down with one sided limits to save on space.

Let’s compute a few more limits using these properties. The next couple of examples will lead us to some truly useful facts about limits that we will use on a continual basis.

Compute the value of the following limit. $$\lim_{x\to -2}(3x^2+5x-9)$$

This first time through we will use only the properties above to compute the limit.

First, we will use property 2 to break up the limit into three separate limits. We will then use property 1 to bring the constants out of the first two limits. Doing this gives us,

$$\begin{array}{cl} \lim\limits_{x\to -2}(3x^2+5x-9) &= \lim\limits_{x\to -2}(3x^2)+\lim\limits_{x\to -2}(5x)-\lim\limits_{x\to -2}(9)\\ &= 3\lim\limits_{x\to -2}(x^2)+5\lim\limits_{x\to -2}(x)-\lim\limits_{x\to -2}(9)\\ \end{array}$$

We can now use properties 7 through 9 to actually compute the limit. $$\begin{array}{cl} \lim\limits_{x\to -2}(3x^2+5x-9) &=3\lim\limits_{x\to -2}(x^2)+5\lim\limits_{x\to -2}(x)-\lim\limits_{x\to -2}(9)\\ &=3(-2)^2+5(-2)-9\\ &=-7\\ \end{array}$$


Now, let’s notice that if we had named this expression $$p(x) = 3x^2 + 5x - 9$$ then the preceding example would have been, $$\begin{array}{cl} \lim\limits_{x\to -2} p(x) &= \lim\limits_{x\to -2}(3x^2+5x-9)\\ &=3(-2)^2+5(-2)-9\\ &=-7\\ &=p(-2)\\ \end{array}$$

In other words, in this case we see that the limit is the same value that we’d get by just evaluating the function at the point in question. This seems to violate one of the main concepts about limits that we’ve seen to this point.

Direct Substitution

In the previous two sections we made a big deal about the fact that limits do not care about what is happening at the point in question. They only care about what is happening around the point. So how does the previous example fit into this since it appears to violate this main idea about limits?

Despite appearances the limit still doesn’t care about what the function is doing at $x=-2$. In this case the function that we’ve got is simply “nice enough” so that what is happening around the point is exactly the same as what is happening at the point. What we mean here by "nice enough" is "continuous." Eventually we will create a calculus definition of continuous; for now we'll depend on the pre-calculus notion of continuity: a function is continuous if it can be drawn without lifting the pencil off the page. That is, it does not have holes, jumps, or vertical asymptotes

The function in the last example was a polynomial. Polynomials are continuous for all real numbers. This means that what is happening around a point at $x=a$ is exactly the same as what is happening at the point where $x=a$. This leads to the following fact.

If $p(x)$ is a polynomial, then, $$\lim_{x\to a}p(x) = p(a)$$


That is, we can compute the limit using direct substitution.


By the end of this section we will generalize this out considerably to most of the functions that we’ll be seeing throughout this course.

Let’s take a look at another example.

Evaluate the following limit
$$\lim_{z\to 1}\frac{6-3z+10z^2}{-2z^4+7z^3+1}$$

Once more, we're greeted with the temptation to use property 4 to begin our work on evaluating the limit. To do so, we must check that $\lim\limits_{z\to 1}(-2z^4+7z^3+1)$ is not 0. The denominator is a polynomial, so we can use direct substitution to evaluate the limit.

$$\lim\limits_{z\to 1}(-2z^4+7z^3+1) = -2(1)^4+7(1)^3+1$$ $$=6$$

Good news, the limit in the denominator will not be 0, so we can use property 4 to evaluate the limit. Therefore,

$$\begin{array}{clcl} \lim\limits_{z\to 1}\frac{6-3z+10z^2}{-2z^4+7z^3+1} &= \frac{\lim\limits_{z\to 1}(6-3z+10z^2)}{\lim\limits_{z\to 1}(-2z^4+7z^3+1)} &~~&\text{Property 4}\\ \end{array}$$

Now, both the numerator and denominator are polynomials so we can use direct substitution to compute the limits of the numerator and the denominator and hence the limit itself.

$$\begin{array}{clcl} \frac{\lim\limits_{z\to 1}(6-3z+10z^2)}{\lim\limits_{z\to 1}(-2z^4+7z^3+1)} &= \frac{6-3(1)+10(1)^2}{-2(1)^4+7(1)^3+1} &~~&\text{direct substitution}\\ &= \frac{13}{6}&&\\ \end{array}$$

In the previous example, as with polynomials, all we really did was evaluate the function at the point in question. So, it appears that there is a fairly large class of functions for which this can be done. Let’s generalize the fact from above a little.

Direct Substitution Property

For functions are said to have the direct substitution property, if $a$ is in the domain of $f(x)$, then,

$$\lim_{x\to a} f(x) = f(a)$$
$$\lim_{x\to a^-} f(x) = f(a)$$
$$\lim_{x\to a^+} f(x) = f(a)$$


As noted in the statement, this fact also holds for the two one-sided limits as well as the normal limit.

Here is a list of some of the more common functions with the direct substitution property.

  • Polynomials .
  • $f(x) = \frac{p(x)}{q(x)}$, for all values of $x$ except where $q(x)=0$.

    • That is, we must exclude $x$-values which make the denominator 0 from the domain of $f(x)$.

  • $\cos(x)$ and $\sin(x)$ .
  • $\tan(x),\sec(x)$.

    • Each of these functions have vertical asymptotes at $x = ...-\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, ...$ etc, so these values of $x$ must be excluded from their domains. For more about these functions, check out this section of our trig textbook.

  • $\cot(x),\csc(x)$ .

    • Each of these functions have vertical asymptotes at $x = ...-\pi, 0, \pi, 2\pi, ...$ etc, so these values of $x$ must be excluded from their domains. For more about these functions, check out this section of our trig textbook.

  • $\sqrt[n]{x}$ for all $x$ if $n$ is odd.
  • $\sqrt[n]{x}$ for $x\geq 0$ if $n$ is even.

    • Here we require $x \geq 0$ to avoid having to deal with complex (imaginary) values.

  • $a^x$ (where $a>0$) and $e^x$ .
  • $\log_b(x)$ (where $b>0$) and $\ln(x)$ for all $x>0$.
  • Any sum, difference or product of the functions on this list will also have the direct substituion property. Quotients will have the direct substitution property on their domains (remembering to check for division by 0, which will create a break in the domain!), as will compositions of functions from this list (also only on their domains).

The last bullet is important. This means that for any combination of these functions all we need to do is evaluate the function at the point in question, making sure that none of the restrictions are violated. This means that we can now do a large number of limits.

Evaluate the following limit.
$$\lim_{x\to 3}\left(-\sqrt[5]{x}+\frac{e^x}{1+\ln(x)}+\sin(x)\cos(x)\right)$$

This is a combination of several of the functions listed above and none of the restrictions are violated so all we need to do is plug in $x=3$ into the function to get the limit.

$$\lim_{x\to 3}\left(-\sqrt[5]{x}+\frac{e^x}{1+\ln(x)}+\sin(x)\cos(x)\right)$$ $$=-\sqrt[5]{3}+\frac{e^3}{1+\ln(3)}+\sin(3)\cos(3)$$ $$\approx 8.1854272743$$

Not a very pretty answer, but we can now calculate the limit.




Practice Problems

  1. Given $\lim\limits_{x\to 8} f(x) = -9, \lim\limits_{x\to 8}g(x) = 2$ and $\lim\limits_{x\to 8}h(x) = 4$, use the limit properties given in this section to compute each of the following limits. If it is not possible to compute any of the limits clearly explain why not.
    1. $\lim\limits_{x\to 8} [3h(x)-6]$
    2. $\lim\limits_{x\to 8}[g(x)h(x)-f(x)]$
    3. $\lim\limits_{x\to 8}[f(x)-g(x)+h(x)]$

    Here is the work for these limits. At each step the property (or properties) used are listed and note that in some cases the properties may have been used more than once in the indicated step.


    1. $\begin{array}{clcl} \lim\limits_{x\to 8} [3h(x)-6] &= \lim\limits_{x\to 8}[3h(x)]-\lim\limits_{x\to 8}(6) &~~&\text{Property 2}\\ &= 3\lim\limits_{x\to 8}h(x) - \lim\limits_{x\to 8}(6) &&\text{Property 1}\\ &= 3\lim\limits_{x\to 8}h(x) - 6 &&\text{Property 7}\\ &= 3(4) -6 &&\text{Substitute in given limit values}\\ &=6 &&\\ \end{array}$


    2. $\begin{array}{clcl} \lim\limits_{x\to 8}[g(x)h(x)-f(x)] &= \lim\limits_{x\to 8}[g(x)h(x)] - \lim\limits_{x\to 8}f(x) &~~&\text{Property 2}\\ &= \left[\lim\limits_{x\to 8}g(x)\right]\left[\lim\limits_{x\to 8}h(x)\right] - \lim\limits_{x\to 8}f(x) &&\text{Property 3}\\ &= (2)(4)-(-9) && \text{Substitute in given limit values}\\ &= 17 &&\\ \end{array}$


    3. $\begin{array}{clcl} \lim\limits_{x\to 8}[f(x)-g(x)+h(x)] &= \lim\limits_{x\to 8}(f(x)) - \lim\limits_{x\to 8}(g(x)) + \lim\limits_{x\to 8}(h(x)) &~~&\text{Property 2}\\ &= -9-2+4 &&\text{Substitute in given limit values}\\ &= -7 &&\\ \end{array}$


  2. Given $\lim\limits_{x\to -4} f(x) = 1, \lim\limits_{x\to -4}g(x) = 10,$ and $\lim\limits_{x\to -4}h(x) = -7$, use the limit properties given in this section to compute each of the following limits. If it is not possible to compute any of the limits clearly explain why not.
    1. $\lim\limits_{x\to -4}\left[\frac{f(x)}{g(x)}-\frac{h(x)}{f(x)}\right]$
    2. $\lim\limits_{x\to -4}[f(x)g(x)h(x)]$
    3. $\lim\limits_{x\to -4} \left[\frac{1}{h(x)}+\frac{3-f(x)}{g(x)+h(x)}\right]$
    4. $\lim\limits_{x\to -4}\left[2h(x)-\frac{1}{h(x)+7f(x)}\right]$

    Here is the work for each limit. At each step the property (or properties) used are listed and note that in some cases the properties may have been used more than once in the indicated step.


    1. $\begin{array}{clcl} \lim\limits_{x\to -4}\left[\frac{f(x)}{g(x)}-\frac{h(x)}{f(x)}\right] &= \lim\limits_{x\to -4}\frac{f(x)}{g(x)} - \lim\limits_{x\to -4}\frac{h(x)}{f(x)} &~~& \text{Property 2}\\ &= \frac{\lim\limits_{x\to -4}f(x)}{\lim\limits_{x\to -4}(g(x)} - \frac{\lim\limits_{x\to -4}h(x)}{\lim\limits_{x\to -4}f(x)} &&\text{Property 4}\\ &= \frac{1}{10}-\frac{-7}{1} &&\text{Substitute in given limit values}\\ &=\frac{71}{10}&&\\ \end{array}$

    2. $\begin{array}{clcl} \lim\limits_{x\to -4}[f(x)g(x)h(x)] &= \left[\lim\limits_{x\to -4}f(x)\right]\left[\lim\limits_{x\to -4}g(x)\right]\left[\lim\limits_{x\to -4}h(x)\right] &~~& \text{Property 3}\\ &= (1)(10)(-7) && \text{Substitute in given limit values}\\ &= -70&&\\ \end{array}$

      3. Note that the properties 2 & 3 in this section were only given with two functions but they can easily be extended out to more than two functions as we did here for property 3.


    3. $\begin{array}{clcl} \lim\limits_{x\to -4}\left[\frac{1}{h(x)}+\frac{3-f(x)}{g(x)+h(x)}\right] &= \lim\limits_{x\to -4}\frac{1}{h(x)} + \lim\limits_{x\to -4}\frac{3-f(x)}{g(x)+h(x)}&~~&\text{Property 2*}\\ &= \frac{\lim\limits_{x\to -4}(1)}{\lim\limits_{x\to -4}h(x)}+\frac{\lim\limits_{x\to -4}[3-f(x)]}{\lim\limits_{x\to -4}[g(x)+h(x)]} && \text{Property 4}\\ &= \frac{\lim\limits_{x\to -4}(1)}{\lim\limits_{x\to -4}h(x)} + \frac{\lim\limits_{x\to -4}(3)-\lim\limits_{x\to -4}f(x)}{\lim\limits_{x\to -4}g(x)+\lim\limits_{x\to -4}h(x)} &&\text{Property 2}\\ &= \frac{1}{-7}+\frac{3-1}{10-7} &&\text{Substitute in given limit values & Property 1}\\ &= \frac{11}{21}&&\\ \end{array}$

      4. *Note that were able to use Property 4 in the second step only because after we evaluated the limit of the denominators (both of them) we found that the limits of the denominators were not zero.


    4. $\begin{array}{clcl} \lim\limits_{x\to -4}\left[2h(x)-\frac{1}{h(x)+7f(x)}\right] &= \lim\limits_{x\to -4}[2h(x)]-\lim\limits_{x\to -4}\frac{1}{h(x)+7f(x)} &~~&\text{Property 2}\\ \end{array}$

      At this point let’s step back a minute. In the previous parts we didn’t worry about using property 4 on a rational expression. However, in this case let’s be a little more careful. We can only use property 4 if the limit of the denominator is not zero. Let’s check that limit and see what we get.

      $\begin{array}{clcl} \lim\limits_{x\to -4} [h(x)+7f(x)] &= \lim\limits_{x\to -4}h(x) + \lim\limits_{x\to -4}[7f(x)] &~~& \text{Property 2}\\ &= \lim\limits_{x\to -4}h(x) + 7\lim\limits_{x\to -4}f(x) && \text{Property 1}\\ &= -7+7(1) &&\text{Substitute in given limit values and Property 1}\\ &= 0&&\\ \end{array}$

      Okay, we can see that the limit of the denominator in the second term will be zero so we cannot actually use property 4 on that term. This means that this limit cannot be done and note that the fact that we could determine a value for the limit of the first term will not change this fact. This limit cannot be done with the given information.


  3. Given $\lim\limits_{x\to 0}f(x) = 6, \lim\limits_{x\to 0}g(x) = -4,$ and $\lim\limits_{x\to 0}h(x) = -1$, use the limit properties given in this section to compute each of the following limits. If it is not possible to compute any of the limits clearly explain why not.
    1. $\lim\limits_{x\to 0}[f(x)+h(x)]^3$
    2. $\lim\limits_{x\to 0}\sqrt{g(x)h(x)}$
    3. $\lim\limits_{x\to 0} \sqrt[3]{11+[g(x)]^2}$
    4. $\lim\limits_{x\to 0} \sqrt{\frac{f(x)}{h(x)-g(x)}}$

    Here is the work for each limit. At each step the property (or properties) used are listed and note that in some cases the properties may have been used more than once in the indicated step.


    1. $\begin{array}{clcl} \lim\limits_{x\to 0}[f(x)+h(x)]^3 &= \left[\lim\limits_{x\to 0}(f(x)+h(x))\right]^3 &~~&\text{Property 5}\\ &= \left[\lim\limits_{x\to 0}f(x) + \lim\limits_{x\to 0}h(x)\right]^3 &&\text{Property 2}\\ &= [6-1]^3 &&\text{Substitute in given limit values}\\ &= 125&&\\ \end{array}$

    2. $\begin{array}{clcl} \lim\limits_{x\to 0}\sqrt{g(x)h(x)} &= \sqrt{\lim\limits_{x\to 0}g(x)h(x)} &~~&\text{Property 6}\\ &= \sqrt{\left[\lim\limits_{x\to 0} g(x)\right]\left[\lim\limits_{x\to 0}h(x)\right]} &&\text{Property 3}\\ &= \sqrt{(-4)(-1)} &&\text{Substitute in given limit values}\\ &= 2&&\\ \end{array}$

    3. $\begin{array}{clcl} \lim\limits_{x\to 0} \sqrt[3]{11+[g(x)]^2} &= \sqrt[3]{\lim\limits_{x\to 0}(11+[g(x)]^2)} &~~&\text{Property 6}\\ &= \sqrt[3]{\lim\limits_{x\to 0}11 + \lim\limits_{x\to 0}[g(x)]^2} &&\text{Property 2}\\ &= \sqrt[3]{\lim\limits_{x\to 0}11 + \left[\lim\limits_{x\to 0}g(x)\right]^2} &&\text{Property 5}\\ &= \sqrt[3]{11+(-4)^2} &&\text{Substitute in given limit values and Property 7}\\ &= 3 &&\\ \end{array}$

    4. $\begin{array}{clcl} \lim\limits_{x\to 0} \sqrt{\frac{f(x)}{h(x)-g(x)}} &= \sqrt{\lim\limits_{x\to 0}\frac{f(x)}{h(x)-g(x)}} &~~& \text{Property 6}\\ &= \sqrt{\frac{\lim\limits_{x\to 0}f(x)}{\lim\limits_{x\to 0}(h(x)-g(x))}} &&\text{Property 4*}\\ &= \sqrt{\frac{\lim\limits_{x\to 0}f(x)}{\lim\limits_{x\to 0}h(x) - \lim\limits_{x\to 0}g(x)}} &&\text{Property 2}\\ &= \sqrt{\frac{6}{-1-(-4)}} && \text{ Substitute in given limit values}\\ &=\sqrt{2} &&\\ \end{array}$

      5. *Note that were able to use Property 4 in the second step only because after we evaluated the limit of the denominators (both of them) we found that the limits of the denominators were not zero.


    For each of the following limits use the limit properties given in this section to compute the limit. At each step clearly indicate the property being used. If it is not possible to compute any of the limits clearly explain why not.

  4. $\lim\limits_{t\to -2}(14-6t+t^3)$
    5. $\begin{array}{clcl} \lim\limits_{t\to -2} (14-6t+t^3) &= \lim\limits_{t\to -2}(14) - \lim\limits_{t\to -2}(6t) + \lim\limits_{t\to -2}(t^3) &~~& \text{Property 2}\\ &= \lim\limits_{t\to -2}(14) - 6 \lim\limits_{t\to -2}(t) + \lim\limits_{t\to -2}(t^3) && \text{Property 1}\\ &= 14 - 6(-2) + (-2)^3 && \text{Properties 7, 8 & 9}\\ &= 18&&\\ \end{array}$

  5. $\lim\limits_{x\to 6}(3x^2+7x-16)$
    6. $\begin{array}{clcl} \lim\limits_{x\to 6}(3x^2+7x-16) &= \lim\limits_{x\to 6}(3x^2) + \lim\limits_{x\to 6}(7x) - \lim\limits_{x\to 6}(16) &~~&\text{Property 2}\\ &= 3 \lim\limits_{x\to 6}(x^2) + 7\lim\limits_{x\to 6}(x) - \lim\limits_{x\to 6}(16) && \text{Property 1}\\ &= 3(6^2) + 7(6) - 16 &&\text{Properties 7, 8, & 9}\\ &= 134 &&\\ \end{array}$

  6. $\lim\limits_{w\to 3}\frac{w^2-8w}{4-7w}$
    7. $\begin{array}{clcl} \lim\limits_{w\to 3}\frac{w^2-8w}{4-7w} &= \frac{\lim\limits_{w\to 3}(w^2-8w)}{\lim\limits_{w\to 3}(4-7w)} &~~& \text{Property 4*}\\ &= \frac{\lim\limits_{w\to 3}(w^2) - \lim\limits_{w\to 3}(8w)}{\lim\limits_{w\to 3}(4)-\lim\limits_{w\to 3}(7w)} &&\text{Property 2}\\ &= \frac{\lim\limits_{w\to 3}(w^2) - 8\lim\limits_{w\to 3}(w)}{\lim\limits_{w\to 3}(4)-7\lim\limits_{w\to 3}(w)} &&\text{Property 1}\\ &= \frac{3^2-8(3)}{4-7(3)} &&\text{Properties 7, 8, & 9}\\ &= \frac{15}{17}&&\\ \end{array}$

    *Note that we were able to use property 4 in the first step because after evaluating the limit in the denominator we found that it wasn’t zero.


  7. $\lim\limits_{x\to -5}\frac{x+5}{x^2+3x-10}$

    8. Okay, at this point let’s step back a minute. We want to use property 4 here and we know that we can only do that if the limit of the denominator is not zero. So, let’s check that out and see what we get.

    $\begin{array}{clcl} \lim\limits_{x\to -5}(x^2+3x-10) &= \lim\limits_{x\to -5}(x^2)+\lim\limits_{x\to -5}(3x)-\lim\limits_{x\to -5}(10) &~~&\text{Property 2}\\ &= \lim\limits_{x\to -5}(x^2) + 3\lim\limits_{x\to -5}(x) - \lim\limits_{x\to -5}(10) && \text{Property 1}\\ &= (-5)^2 + 3(-5) - 10 &&\text{Properties 7, 8, and 9}\\ &= 0 &&\\ \end{array}$

    So, the limit of the denominator is zero so we couldn’t use property 4 in this case. Therefore, we cannot do this limit at this point (note that it will be possible to do this limit after the next section).


  8. $\lim\limits_{z\to 0}\sqrt{z^2+6}$
    9. $\begin{array}{clcl} \lim\limits_{z\to 0}\sqrt{z^2+6} &= \sqrt{\lim\limits_{z\to 0}(z^2+6)} &~~&\text{Property 6}\\ &= \sqrt{\lim\limits_{z\to 0}(z^2) + \lim\limits_{z\to 0}(6)} &&\text{Property 2}\\ &= \sqrt{0^2+6} &&\text{Properties 7, 8, and 9}\\ &= \sqrt{6} &&\\ \end{array}$

  9. $\lim\limits_{x\to 10}(4x+\sqrt[3]{x-2})$
    10. $\begin{array}{clcl} \lim\limits_{x\to 10}(4x+\sqrt[3]{x-2}) &= \lim\limits_{x\to 10}(4x) + \lim\limits_{x\to 10}\sqrt[3]{x-2} &~~&\text{Property 2}\\ &= \lim\limits_{x\to 10}(4x) + \sqrt[3]{\lim\limits_{x\to 10}(x-2)} && \text{Property 6}\\ &= \lim\limits_{x\to 10}(4x) + \sqrt[3]{\lim\limits_{x\to 10}(x) - \lim\limits_{x\to 10}(2)} &&\text{Property 2}\\ &= 4\lim\limits_{x\to 10}(x) + \sqrt[3]{\lim\limits_{x\to 10}(x) - \lim\limits_{x\to 10}(2)} &&\text{Property 1}\\ &= 4(10) + \sqrt[3]{10-2} && \text{Properties 7 & 8}\\ &= 42 &&\\ \end{array}$

Assignment Problems

  1. Given $\lim\limits_{x\to 0}f(x) = 5, \lim\limits_{x\to 0}g(x) = -1$, and $\lim\limits_{x\to 0}h(x) = -3$, use the limit properties given in this section to compute each of the following limits. If it is not possible to compute any of the limits clearly explain why not.
    1. $\lim\limits_{x\to 0}[11+7f(x)]$
    2. $\lim\limits_{x\to 0}[6-4g(x)-10h(x)]$
    3. $\lim\limits_{x\to 0}[4g(x) -12f(x) + 3h(x)]$
    4. $\lim\limits_{x\to 0}[g(x)(1+2f(x))]$

  2. Given $\lim\limits_{x\to 12}f(x) = 2, \lim\limits_{x\to 12}g(x) = 6$ and $\lim\limits_{x\to 12}h(x) = 9$, use the limit properties given in this section to compute each of the following limits. If it is not possible to compute any of the limits clearly explain why not.
    1. $\lim\limits_{x\to 12}\left[h(x)f(x) +\frac{1+g(x)}{g(x)}\right]$
    2. $\lim\limits_{x\to 12}[(3-f(x))(1+2g(x))]$
    3. $\lim\limits_{x\to 12}\frac{f(x)+1}{3g(x)-2h(x)}$
    4. $\lim\limits_{x\to 12}\frac{f(x)-2g(x)}{7+h(x)f(x)}$

  3. Given $\lim\limits_{x\to -1}f(x) = 0, \lim\limits_{x\to -1}g(x) = 9,$ and $\lim\limits_{x\to -1}h(x) = -7$, use the limit properties given in this section to compute each of the following limits. If it is not possible to compute any of the limits clearly explain why not.
    1. $\lim\limits_{x\to -1}[(g(x))^2-(h(x))^3]$
    2. $\lim\limits_{x\to -1}\sqrt{3+6f(x)-h(x)}$
    3. $\lim\limits_{x\to -1}\sqrt{f(x)-g(x)h(x)}$
    4. $\lim\limits_{x\to -1}\sqrt[4]{\frac{2+g(x)}{1-10h(x)}}$

For each of the following limits use the limit properties given in this section to compute the limit. At each step clearly indicate the property being used. If it is not possible to compute any of the limits clearly explain why not.

  1. $\lim\limits_{x\to 4}(3x^2-9x+2)$
  2. $\lim\limits_{w\to -1}(w-(w^2+3)^2)$
  3. $\lim\limits_{t\to 0}(t^4-4t^2+12t-8)$
  4. $\lim\limits_{z\to 2}\frac{10+z^2}{3-4z}$
  5. $\lim\limits_{x\to 7}\frac{8x}{x^2-14x+49}$
  6. $\lim\limits_{y\to -3}\frac{y^3-20y+4}{y^2+8y-1}$
  7. $\lim\limits_{w\to -6}\sqrt[3]{8+7w}$
  8. $\lim\limits_{t\to 1}(4t^2-\sqrt{8t+1})$
  9. $\lim\limits_{x\to 8}(\sqrt[4]{3x-8}+\sqrt{9+2x})$

Note: The following content is adapted from OpenStax and is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction.

In the following exercises, use the following graphs and the limit laws to evaluate each limit.

f(x), which has two linear segments. The first is a line with negative slope existing for x < -3. It goes toward the point (-3,0) at x= -3. The next has increasing slope and goes to the point (-3,-2) at x=-3. It exists for x > -3. Other key points are (0, 1), (-5,2), (1,2), (-7, 4), and (-9,6).
g(x), which has a linear segment and a curved segment. The linear segment exists for x < -3 and has decreasing slope. It goes to (-3,-2) at x=-3. The curved segment appears to be the right half of a downward opening parabola. It goes to the vertex point (-3,2) at x=-3. It crosses the y axis a little below y=-2. Other key points are (0, -7/3), (-5,0), (1,-5), (-7, 2), and (-9, 4).
  1. $\lim\limits_{x\to-3^+}(f(x)+g(x))$
  2. $\lim\limits_{x\to-3^-}(f(x)-3g(x))$
  3. $\lim\limits_{x\to 0}\frac{f(x)g(x)}{3}$
  4. $\lim\limits_{x\to -5}\frac{2+g(x)}{f(x)}$
  5. $\lim\limits_{x\to1}(f(x))^2$
  6. $\lim\limits_{x\to1}\sqrt[3]{f(x)-g(x)}$
  7. $\lim\limits_{x\to-7}(x\cdot g(x))$
  8. $\lim\limits_{x\to-9}[x\cdot f(x) + 2\cdot g(x)]$


  9. The density of an object is given by its mass divided by its volume: $\rho = \frac{m}{V}$.
    1. Use a calculator to plot the volume as a function of density ($V=m/\rho$),assuming you are examining something of mass 8 kg ( $m=8$).
    2. Evaluate $\lim\limits_{\rho\to0^+}V(\rho)$ and explain the physical meaning.

    Indicate whether the following statements are true or false. Justify your answers.

  10. $\lim\limits_{x\to -5}\frac{x+5}{x-2} = \frac{\lim\limits_{x\to -5}(x+5)}{\lim\limits_{x\to -5}(x-2)}$
  11. $\lim\limits_{x\to 2}\frac{x+5}{x-2} = \frac{\lim\limits_{x\to 2}(x+5)}{\lim\limits_{x\to 2}(x-2)}$
  12. $\lim\limits_{x\to a}f(x) = f(a)$ for all functions $f(x)$.
  13. If $\lim\limits_{x\to 3}f(x) = 7$ and $\lim\limits_{x\to 3}g(x) = -2$, then $\lim\limits_{x\to 3}\frac{f(x)}{g(x)+2}$ does not exist.
  14. If $\lim\limits_{x\to 3}f(x) = 7$ and $\lim\limits_{x\to 3}g(x) = -2$, then $\lim\limits_{x\to 3}\frac{7-f(x)}{g(x)}$ does not exist.