Limits Summary

As we do for all limits, let's start by attempting direct substitution. Since $x$ is going to $\infty$, this means instead reasoning about infinities using some of our limit properties. \[\lim\limits_{x\to\infty}\dfrac{3x^2+5x}{x^2+2x+1}\Rightarrow\dfrac{3(\infty)^2+5(\infty)}{(\infty)^2+2(\infty)+1} \Rightarrow \dfrac{\infty}{\infty}\] Here we get an indeterminate form $\infty/\infty$, so our goal is to use algebra to simplify this so we don't get $\infty/\infty$ (see Limits at Infinity part I). If we ask ourselves what is causing the $\infty$ on bottom and top, we realize that it is from the largest power of $x$, namely $x^2$. For this reason, we are going to factor out the portion of the numerator and denominator causing the indeterminate form, which happens to be $x^2$ for both.

Now we reasoning with infinities again and see what the new situation is.

This time, we got a value for the limit, so we are done and can conclude $\lim\limits_{x\to\infty}\dfrac{3x^2+5x}{x^2+2x+1} = 3$.

Again, let's start by trying direct substitution. \[\lim\limits_{x\to-1}\dfrac{3x^2+5x}{x^2+2x+1} \Rightarrow \dfrac{3(-1)^2+5(-1)}{(-1)^2+2(-1)+1} \Rightarrow \dfrac{-2}{0}\] This time, we got a nonzero number divided by $0$. This is not indeterminate, but rather an indication that the fraction is going towards $\pm \infty$ (see the infinite limits section). Our next question is whether this is fraction is always positive or always negative as $x$ approaches $-1$ with the limit. If it is both, then the limit does not exist. If we can show its always positive, for example, then we could say that the value of the limit goes towards $\infty$.

For the numerator, as $x$ approaches $-1$, we saw from direct substitution that the numerator is approaching $-2$. This is a negative number. For the denominator, since we get $0$, let's factor it and see if that can make the sign more clear. \[x^2+2x+1 = (x+1)^2\] From this, we see that the denominator is always squared, and so it is positive (except at $x = -1$, where the denominator becomes $0$). Thus the denominator is approaching $0$ with positive values. We can write the work like this: \[\lim\limits_{x\to-1}\dfrac{3x^2+5x}{x^2+2x+1} = \lim\limits_{x\to-1}\dfrac{3x^2+5x}{(x+1)^2}\Rightarrow \dfrac{-2}{0^+} \Rightarrow -\infty\] This shows that $\lim\limits_{x\to-1}\dfrac{3x^2+5x}{x^2+2x+1} = -\infty$.

We knew to apply techniques from the infinite limits section because direct substitution initially showed us a nonzero number divided by zero.

As usual, we start with direct substitution. \[\lim\limits_{x\to0}\dfrac{3x^2+5x}{x^2+2x+1} = \dfrac{3(0)^2+5(0)}{(0)^2+2(0)+1} = \dfrac{0}{1} = 0\] Direct substitution gave us an immediate solution, and so we don't need any other techniques, and we have already found the value of the limit to be $0$.

Let's start with direct substitution.\[\lim\limits_{x\to3}\dfrac{3x^3-9x^2}{x^2+4x-21} \Rightarrow \dfrac{81-81}{9+12-21} \Rightarrow \dfrac00\] It probably doesn't surprise you that this is $0/0$ because that's what we're reviewing right now, but we generally won't have this context when evaluating limits, so it is good practice!

Since $x$ is approaching $3$ and the limit has the indeterminate form $0/0$, our goal is to try and factor out $x-3$ if possible from the numerator and denominator. As a brief aside, the fundamental theorem of algebra guarantees this is possible for the numerator and denominator separately, as it says that if $p(a) = 0$ for any polynomial $p(x)$, then $(x-a)$ can be factored out of $p(x)$. So let's try to factor the numerator and denominators separately, expecting $x-3$ as one of the factors. \[\dfrac{3x^3-9x^2}{x^2+4x-21} = \dfrac{3x^2(x-3)}{(x-3)(x+7)}\]

With this algebra, we can now cancel the factors of $x-3$ and try direct substitution again. $$\begin{align*} \lim\limits_{x\to3}\dfrac{3x^3-9x^2}{x^2+4x-21} &= \lim\limits_{x\to3}\dfrac{3x^2(x-3)}{(x-3)(x+7)}\\ &= \lim\limits_{x\to3}\dfrac{3x^2}{x+7} \\ &= \dfrac{3(3)^2}{3+7} \\ &= \dfrac{27}{10} \end{align*}$$ Thus $\lim\limits_{x\to3}\dfrac{3x^3-9x^2}{x^2+4x-21} = \dfrac{27}{10}$.

Let's try direct substitution. \[\lim\limits_{x\to4}\dfrac{4-x}{\sqrt{2x+1}-3} \Rightarrow \dfrac{4-4}{\sqrt{2(4)+1}-3} \Rightarrow \dfrac{0}{0}\] This time, the denominator is not a polynomial, so we can't simply try to factor out $x-4$. We can factor $x-4$ out of the numerator though, and hope to get an $x-4$ to cancel with it from multiplying by the conjugate in the denominator. Because of this, we will keep the $x-4$ we pull out of the numerator to the side and NOT distribute it when we multiply by the conjugate.

It worked! Multiplying by the conjugate allowed us to get an $(x-4)$ in the denominator that we can now cancel with the $x-4$ we factored out of the numerator. Once we cancel these portions, we'll try direct substitution again.

Thus $\lim\limits_{x\to4}\dfrac{4-x}{\sqrt{2x+1}-3} = -3$.

Let's try the equivalent of direct substitution for infinite limits. \[\lim\limits_{x\to-\infty}\dfrac{3x+5}{8x^2+3x+1} \Rightarrow \dfrac{3(-\infty)+5}{8(-\infty)^2+3(-\infty)+1} \Rightarrow \dfrac{-\infty}{\infty - \infty}\]

This is an indeterminate form involving infinity and a fraction. So let's factor the highest powers of $x$ out of the denominator and numerator separately and cancel them as much possible. Then we'll try reasoning with infinities again.

Thus $\lim\limits_{x\to-\infty}\dfrac{3x+5}{8x^2+3x+1} = 0$.

Reasoning with infinities for this limits yields \[\lim\limits_{x\to\infty}\dfrac{3+2e^x}{7-4e^x} \Rightarrow \dfrac{3+e^\infty}{7-4e^\infty} \Rightarrow \dfrac{3+\infty}{7-4(\infty)} \Rightarrow \dfrac{\infty}{\infty}\]

This time, we don't have polynomials. But hopefully it isn't that big a stretch to see that it is the $e^x$ that is causing $\infty$ in both the numerator and denominator as $x$ goes to $\infty$. So let's try factoring out $e^x$ from each and then cancelling those from the top and bottom.

Thus $\lim\limits_{x\to\infty}\dfrac{3+2e^x}{7-4e^x} = -\dfrac{1}{2}$.

Let's start with reasoning with infinities. \[\lim\limits_{x\to-\infty}\dfrac{3+2e^x}{7-4e^x} \Rightarrow \dfrac{3+2e^{-\infty}}{7-4e^{-\infty}} \Rightarrow \dfrac{3+2(0)}{7-4(0)} = \dfrac37\] This time, there was no indeterminate form, and reasoning with infinities showed us almost immediately (after using properties of limits involving exponential functions and infinity) that $\lim\limits_{x\to-\infty}\dfrac{3+2e^x}{7-4e^x} = \dfrac37$

Reasoning with infinities for this limit shows us \[\lim\limits_{x\to\infty} (2x^9 - 3x^5) \Rightarrow 2(\infty)^9 - 3(\infty)^5 \Rightarrow \infty - \infty\]

This time, we do get an indeterminate form. This is just a polynomial, not a rational function, so let's just factor out the largest power of $x$ and then try using properties of infinity again.

Thus $\lim\limits_{x\to\infty} (2x^9 - 3x^5) = \infty$.

Let's try direct substitution. \[\lim\limits_{x\to5^-}\dfrac{2x-1}{25-x^2} \Rightarrow \dfrac{2(5)-1}{25-(5)^2} \Rightarrow \dfrac{9}{0}\]

Since we get a nonzero number divided by zero, we know the value of the limit is $\infty, -\infty,$ or DNE. We know the numerator is going to $9$, let's see if factoring can help us with the denominator. \[\lim\limits_{x\to5^-}\dfrac{2x-1}{25-x^2} = \lim\limits_{x\to5^-} \dfrac{2x-1}{(5-x)(5+x)} \Rightarrow \dfrac{9}{(0^+)(10)} \Rightarrow \infty\]

Note that we knew $(5-x)$ was approaching $0$ from the right side because $x$ is approaching $5$ from its left side, and hence $5$ is bigger than $x$. Putting this all together gives us the final result $\lim\limits_{x\to5^-}\dfrac{2x-1}{25-x^2} = \infty$.

This is nearly the same as the previous limit, but this time the limit is from both sides. Therefore, it is no surprise that we get the same initial result from direct substitution: \[\lim\limits_{x\to5^-}\dfrac{2x-1}{25-x^2} \Rightarrow \dfrac{2(5)-1}{25-(5)^2} \Rightarrow \dfrac{9}{0}\]

Since the limit is approaching from both sides, and we've already shown $\lim\limits_{x\to5^-}\dfrac{2x-1}{25-x^2} = \infty$, let's check the other side of the limit using the same factoring as before. \[\lim\limits_{x\to5^+}\dfrac{2x-1}{25-x^2} = \lim\limits_{x\to5^+} \dfrac{2x-1}{(5-x)(5+x)} \Rightarrow \dfrac{9}{(0^-)(10)} \Rightarrow -\infty\]

Note that we knew $(5-x)$ was approaching $0$ from the left side because $x$ is approaching $5$ from its left side, and hence $x$ is bigger than $5$.

Since the one-sided limits yield different values of infinity, the whole limit does not exist.

If we try direct substitution, we get \[\lim\limits_{x\to0}\dfrac{9x+7}{x^3-5x^2} \Rightarrow \dfrac{7}{0}\]

Since we get a nonzero number divided by zero, we know the value of the limit is $\infty, -\infty,$ or DNE. We know the numerator is going to $7$, let's see if factoring can help us with the denominator. \[\lim\limits_{x\to0}\dfrac{9x+7}{x^5-5x^4} = \lim\limits_{x\to0}\dfrac{9x+7}{x^4(x-5)} \Rightarrow \dfrac{7}{(0^+)(-5)}\Rightarrow -\infty\]

In this case, we knew $x^4$ was approaching $0$ from the right side because any real number to an even power is positive, and hence $x^4$ is positive as $x$ approaches $0$ (from both sides of $0$). Thus $\lim\limits_{x\to0}\dfrac{9x+7}{x^5-5x^4} = -\infty$.

If we try direct substitution, we get \[\lim\limits_{x\to3} |x-3|\sin\left(\dfrac2{6-2x}\right) \Rightarrow 0\sin\left(\dfrac20\right)\]

Although we could test if the $\frac20$ is going to $\infty$ or $-\infty$, it wouldn't matter since we can't plug $\infty$ or $-\infty$ into sine. Instead, we will try to use the Squeeze Theorem to obtain some simpler limits to evaluate.

This limit involves sine, so our starting place is that the output of sine is between $-1$ and $1$. Let's write this down with the specific input sine has in the limit: \[-1 \leq \sin\left(\dfrac2{6-2x}\right) \leq 1\]

But the function in our limit isn't simply $\sin(\frac2{6-2x})$, it's $|x-3|\sin(\frac2{6-2x})$. To get to the function we need, we'll multiply the above inequality through by $|x-3|$. Note that $|x-3| \geq 0$, so the direction of the inequalities will remain consistent. \[-|x-3| \leq |x-3|\sin\left(\dfrac2{6-2x}\right) \leq |x-3|\]

This gives us the upper and lower functions we can use for the Squeeze Theorem, namely $f(x) = -|x-3|$ and $h(x) = |x-3|$, which look much simpler than the function in the limit $g(x) = |x-3|\sin(\frac2{6-2x})$. All that's left before we can use the conclusion of the Squeeze Theorem is to check the limits as $x$ approaches $3$ of both $f(x)$ and $g(x)$. \[\lim\limits_{x\to3} -|x-3| = 0 \hspace{3em} \lim\limits_{x\to3} |x-3| = 0\]

We evaluated both limits with direct substitution, and both limits are going to $0$, so by the Squeeze Theorem, we conclude $\lim\limits_{x\to3} |x-3|\sin\left(\dfrac2{6-2x}\right) = 0$.

Unfortunately, without a formula for $p(t)$, we cannot even attempt direct substitution for the limit. Since we know some information about how large and small the value of $p(t)$ can get, we'll try to use the Squeeze Theorem. We know \[2 \leq p(t) \leq 6\]

This is our starting place, and from here we are going to try and construct the function in the limit $t^2p(t)+5$. To do this, we'll simply multiply through the inequality by $t^2$ and then add $5$ on all sides to get \[2t^2+5 \leq t^2p(t)+5 \leq 6t^2+5\]

From this, we can see the lower, middle, and upper functions for the Squeeze Theorem are $f(t) = 2t^2+5$, $g(t) = t^2p(t)+5$, and $h(t) = 6t^2+5$. In particular, $f(t)$ and $h(t)$ have explicit formulas where we can attempt direct substitution. So let's see what their limits evaluate to as $t$ approaches $0$.

Thus by the Squeeze Theorem we conclude $\lim\limits_{t\to0} (t^2p(t)+5) = 5$, even without knowing any specific formula for $p(t)$.