Proofs of Limit Properties
In this section we are going to prove some of the basic properties and facts about limits that we saw in the Limits chapter. Before proceeding with any of the proofs we should note that many of the proofs use the precise definition of the limit and it is assumed that not only have you read that section but that you have a fairly good feel for doing that kind of proof. If you’re not very comfortable using the definition of the limit to prove limits you’ll find many of the proofs in this section difficult to follow.
The proofs that we’ll be doing here will not be quite as detailed as those in the precise definition of the limit section. The “proofs” that we did in that section first did some work to get a guess for the $\delta$ and then we verified the guess. The reality is that often the work to get the guess is not shown and the guess for $\delta$ is just written down and then verified. For the proofs in this section where a $\delta$ is actually chosen we’ll do it that way. To make matters worse, in some of the proofs in this section work very differently from those that were in the limit definition section.
So, with that out of the way, let’s get to the proofs.
Limit Properties
In the Limit Properties section we gave several properties of limits. We’ll prove most of them here. First, let’s recall the properties here so we have them in front of us. We’ll also be making a small change to the notation to make the proofs go a little easier.
Note that we added values ($K, L$, etc.) to each of the limits to make the proofs much easier. In these proofs we’ll be using the fact that we know $\lim\limits_{x\to a}f(x) = K$ and $\lim\limits_{x\to a}g(x) = L$ and we’ll use the definition of the limit to make a statement about $|f(x)-K|$ and $|g(x)-L|$ which will then be used to prove what we actually want to prove. When you see these statements do not worry too much about why we chose them as we did. The reason will become apparent once the proof is done.
Also, we’re not going to be doing the proofs in the order they were first presented in the Limits chapter. Some of the proofs will be easier if we’ve got some of the others proved first.
Proof: To make the notation a little clearer let’s define the function $f(x) = c$. Then what we're being asked to prove is that $\lim\limits_{x\to a}f(x) = c.$
Let $\epsilon \gt 0$; we need to show that we can find a $\delta \gt 0$ so that $$|f(x) - c| \lt \epsilon ~~~~~\text{whenever}~~~~~ 0 \lt |x-a| \lt \delta.$$
The left inequality is trivially satisfied for any $x$, however, because we defined $f(x) = c$. So simply choose $\delta \gt 0$ to be any number you want (you generally can’t do this with these proofs). Then, $$|f(x) - c| = |c-c| = 0 \lt \epsilon.$$
Proof: There are several ways to prove this part. If you accept 3 And 7 then all you need to do is let $g(x) = c$ and then this is a direct result of 3 and 7. However, we’d like to do a more rigorous mathematical proof. So here is that proof.
Case 1, $c=0$: First, note that if $c=0$ then $cf(x) = 0$ and so, $$\lim_{x\to a}[0f(x) = \lim_{x\to a}(0) = 0 = 0f(x).$$
The limit evaluation here is a special case of 7 (with $c=0$), which we just proved. Therefore we know 1 is true for $c=0$, and so we can assume that $c \neq 0$ for the remainder of the proof.
First, note that we’ll need something called the triangle inequality in this proof. The triangle inequality states that, $$|a+b|\leq |a|+|b|.$$
Proof: We’ll be doing this proof in two parts. First let’s prove $\lim\limits_{x\to a}[f(x)+g(x)] = K+L$.
Let $\epsilon \gt 0$. Then because $\lim\limits_{x\to a}f(x) = K$ and $\lim\limits_{x\to a}g(x) = L$, there are a $\delta_1\gt 0$ and a $\delta_2 \gt 0$ such that, $$|f(x)-K|\lt \frac{\epsilon}{2}~~~~~ \text{whenever}~~~~~ 0\lt |x-a| \lt \delta_1$$ $$|g(x)-L|\lt \frac{\epsilon}{2}~~~~~ \text{whenever}~~~~~ 0\lt |x-a| \lt \delta_2.$$
Now, we can choose $\delta = \min\{\delta_1, \delta_2\}$. Then we need to show that $$|f(x) + g(x) - (K+L)|\lt \epsilon ~~~~~ \text{whenever}~~~~~ 0 \lt |x-a| \lt \delta.$$
If $0 \lt |x-a|\lt \delta$, we then have, $$\begin{align*} |f(x) + g(x) - (K+L)| &= |(f(x)-K)+(g(x)-L)| \\ &\leq |f(x) - K|+|g(x)-L| ~~\text{by the triangle inequality}\\ &\lt \frac{\epsilon}{2}+\frac{\epsilon}{2}\\ &= \epsilon\\ \end{align*}$$
This one is a little tricky. First, let's note that because $\lim\limits_{x\to a}f(x) = K$ and $\lim\limits_{x\to a}g(x) = L$, we can use properties 2 and 7 to prove the following two limits.
$$\lim_{x\to a}[f(x)-K] = \lim_{x\to a}f(x) - \lim_{x \to a}K = K-K= 0$$ $$\lim_{x\to a}[g(x)-L] = \lim_{x\to a}g(x) - \lim_{x \to a}L = L-L= 0$$Now, let $\epsilon \gt 0$. Then there is a $\delta_1 \gt 0$ and a $\delta_2 \gt 0$ such that $$|(f(x)-K) - 0| \lt \sqrt{\epsilon}~~~~\text{whenever}~~~~ 0 \lt |x-a| \lt \delta_1$$ $$|(g(x)-L) - 0| \lt \sqrt{\epsilon}~~~~\text{whenever}~~~~ 0 \lt |x-a| \lt \delta_2.$$
Choose $\delta = \min\{\delta_1, \delta_2\}$. If $0 \lt |x-a|\lt \delta$, we then get, $$|[f(x) - K][g(x) - L]-0| = |f(x) - K||g(x) -L|$$ $$\lt \sqrt{\epsilon}\sqrt{\epsilon}$$ $$=\epsilon.$$
So, we've managed to prove that $$\lim_{x\to a}[f(x) - K][g(x)-L] = 0.$$
This apparently has nothing to do with what we actually want to prove, but as you’ll see in a bit it is needed.
Before launching into the actual proof of 3 let’s do a little Algebra. First, expand the following product.
$$[f(x)-K][g(x)-L] = f(x)g(x) - Lf(x)-Kg(x)+KL$$Rearranging this gives the following way to write the product of the two functions.
$$f(x)g(x) = [f(x)-K][g(x)-L]+Lf(x)+Kg(x)-KL$$With this we can now proceed with the proof of property 3.
$$\begin{align*}\mathop {\lim }\limits_{x \to a} f\left( x \right)g\left( x \right) & = \mathop {\lim }\limits_{x \to a} \left[ {\left[ {f\left( x \right) - K} \right]\left[ {g\left( x \right) - L} \right] + Lf\left( x \right) + Kg\left( x \right) - KL} \right]\\ & = \mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - K} \right]\left[ {g\left( x \right) - L} \right] + \mathop {\lim }\limits_{x \to a} Lf\left( x \right) + \mathop {\lim }\limits_{x \to a} Kg\left( x \right) - \mathop {\lim }\limits_{x \to a} KL\\ & = 0 + \mathop {\lim }\limits_{x \to a} Lf\left( x \right) + \mathop {\lim }\limits_{x \to a} Kg\left( x \right) - \mathop {\lim }\limits_{x \to a} KL\hspace{0.5in}\hspace{0.5in}\\ & = LK + KL - KL\\ & = LK\end{align*}$$Fairly simple proof really, once you see all the steps that you have to take before you even start. The second step made multiple uses of property 2. In the third step we used the limit we initially proved. In the fourth step we used properties 1 and 7. Finally, we just did some simplification.
This one is also a little tricky. First, we’ll start of by proving, $$\lim_{x\to a}\frac{1}{g(x)}=\frac{1}{L}.$$
Let $\epsilon \gt 0$. We’ll not need this right away, but these proofs always start off with this statement. Now, because $\lim\limits_{x\to a}g(x) = L$, there is a $\delta_1\gt 0$ such that $$|g(x) - L|\lt \frac{|L|}{2} ~~~~ \text{whenever} ~~~~ 0 \lt |x-a| \lt \delta_1.$$
Now, assuming that $0\lt |x-a| \lt \delta_1$, we have, $$\begin{align*}\left| L \right| & = \left| {L - g\left( x \right) + g\left( x \right)} \right| & \hspace{0.5in} & {\mbox{just adding zero to }}L\\ & \le \left| {L - g\left( x \right)} \right| + \left| {g\left( x \right)} \right| & \hspace{0.5in} & {\mbox{using the triangle inequality}}\\ & = \left| {g\left( x \right) - L} \right| + \left| {g\left( x \right)} \right| & \hspace{0.5in} & \left| {L - g\left( x \right)} \right| = \left| {g\left( x \right) - L} \right|\\ & < \frac{{\left| L \right|}}{2} + \left| {g\left( x \right)} \right| & \hspace{0.5in} & {\mbox{assuming that }}0 < \left| {x - a} \right| < {\delta _1}\end{align*}$$ Rearranging this gives, $$\left| L \right| < \frac{{\left| L \right|}}{2} + \left| {g\left( x \right)} \right|\hspace{0.25in} \Rightarrow \hspace{0.25in}\frac{{\left| L \right|}}{2} < \left| {g\left( x \right)} \right|\hspace{0.5in} \Rightarrow \hspace{0.25in}\frac{1}{{\left| {g\left( x \right)} \right|}} < \frac{2}{{\left| L \right|}}.$$
Now, there is also a $\delta_2 \gt 0$ such that $$\left| {g\left( x \right) - L} \right| < \frac{{{{\left| L \right|}^2}}}{2}\varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < {\delta _{\,2}}.$$
Choose $\delta = \min\{\delta_1, \delta_2\}$. If $0 \lt |x-a|\lt \delta$ we have, $$\begin{align*}\left| {\frac{1}{{g\left( x \right)}} - \frac{1}{L}} \right| & = \left| {\frac{{L - g\left( x \right)}}{{Lg\left( x \right)}}} \right| & \hspace{0.5in} & {\mbox{common denominators}}\\ & = \frac{1}{{\left| {Lg\left( x \right)} \right|}}\left| {L - g\left( x \right)} \right| & \hspace{0.25in} & {\mbox{doing a little rewriting}}\\ & = \frac{1}{{\left| L \right|}}\frac{1}{{\left| {g\left( x \right)} \right|}}\left| {g\left( x \right) - L} \right| & \hspace{0.25in} & {\mbox{doing a little more rewriting}}\\ & < \frac{1}{{\left| L \right|}}\frac{2}{{\left| L \right|}}\left| {g\left( x \right) - L} \right| & \hspace{0.25in} & {\mbox{assuming that }}0 < \left| {x - a} \right| < \delta \le {\delta _1}\\ & < \frac{2}{{{{\left| L \right|}^2}}}\frac{{{{\left| L \right|}^2}}}{2}\varepsilon & \hspace{0.5in} & {\mbox{assuming that }}0 < \left| {x - a} \right| < \delta \le {\delta _2}\\ & = \varepsilon \end{align*}$$
Now that we've proven $\lim\limits_{x\to a}\frac{1}{g(x)}=\frac{1}{L}$, the more general fact is easy.
$$\begin{align*}\mathop {\lim }\limits_{x \to a} \left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] & = \mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right)\frac{1}{{g\left( x \right)}}} \right]\\ & = \mathop {\lim }\limits_{x \to a} f\left( x \right)\mathop {\lim }\limits_{x \to a} \frac{1}{{g\left( x \right)}}\hspace{0.5in}{\mbox{using property }}{\bf{3}}{\mbox{.}}\\ & = K\frac{1}{L} = \frac{K}{L}\end{align*}$$Proof for $n$ an integer: As noted we’re only going to prove 5 for integer exponents. This will also involve proof by induction so if you aren’t familiar with induction proofs you can skip this proof.
So, we’re going to prove, $$\mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^n} = {\left[ {\mathop {\lim }\limits_{x \to a} f\left( x \right)} \right]^n} = {K^n},\hspace{0.5in}n \ge 2,\,\,n{\mbox{ is an integer}}{\mbox{.}}$$
For $n = 2$ we have nothing more than a special case of property 3.
$$\mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^2} = \mathop {\lim }\limits_{x \to a} f\left( x \right)f\left( x \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right)\mathop {\lim }\limits_{x \to a} f\left( x \right) = KK = {K^2}$$So the property is proven for $n = 2$. Now assume that 5 is true for $n-1$, or $\lim\limits_{x\toa}[f(x)]^{n-1} = K^{n-1}$. Then, again using property 3 we have $$\begin{align*}\mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^n} & = \mathop {\lim }\limits_{x \to a} \left( {{{\left[ {f\left( x \right)} \right]}^{n - 1}}f\left( x \right)} \right)\\ & = \mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^{n - 1}}\mathop {\lim }\limits_{x \to a} f\left( x \right)\\ & = {K^{n - 1}}K\\ & = {K^n}\end{align*}$$
As pointed out in the Limit Properties section this is nothing more than a special case of the full version of 5 and the proof is given there and so is the proof is not give here.
This is a simple proof. If we define $f(x) = x$ to make the notation a little easier, we're being asked to prove that $\lim\limits_{x\to a}f(x) = a$.
Let $\epsilon \gt 0$ and let $\delta = \epsilon$. Then if $0 \lt |x-a| \lt \delta = \epsilon$ we have, $$|f(x) - a| = |x-a| \lt \delta = \epsilon.$$ Therefore we've proved that $\lim\limits_{x\to a}(x) = a$.
This is just a special case of property 5 with $f(x) = x$ and so we won’t prove it here.
Facts, Infinite Limits
Given the functions $f(x)$ and $g(x)$, suppose we have $$\lim_{x\to c}f(x) = \infty ~~~~~ \lim_{x\to c}g(x) = L$$ for some real numbers $c$ and $L$.
Fact 1: $\lim\limits_{x\to c}[f(x)\pm g(x)] = \infty$.Partial proof of 1: We will prove $\lim\limits_{x\to c}[f(x)+g(x)]=\infty$ here. The proof of $\lim\limits_{x\to c}[f(x)-g(x)]=\infty$ is nearly identical and is left to you.
Let $M \gt 0$. Then because we know $\lim\limits_{x\to c}f(x) = \infty$, there exists a $\delta_1 \gt 0$ such that if $0 \lt |x-c|\lt \delta_1$, we have, $$f(x) \gt M-L+1.$$
Also, because we know $\lim\limits_{x\to c}g(x) = L$, there exists a $\delta_2 \gt 0$ such that if $0 \lt |x-c| \lt \delta_2$, we have, $$0 < \left| {g\left( x \right) - L} \right| < 1\hspace{0.25in}\,\, \to \hspace{0.25in} - 1 < g\left( x \right) - L < 1\hspace{0.25in} \to \,\,\,\,\,\,\,L - 1 < g\left( x \right) < L + 1.$$
Now, let $\delta = \min\{\delta_1, \delta_2\}$, and so if $0 \lt |x-c|\lt \delta$, we know from the above statements that we will have both, $$f(x) \gt M-L+1~~~~~ g9x) \gt > - 1$$