Change of Variables
Motivation
Recall that $\displaystyle \lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$. Now, suppose we want to evaluate $\displaystyle \lim_{x \rightarrow 0} \frac{\sin(kx)}{x}$ for some constant $k \neq 0$. How can we do this?
Notice, we can rewrite the limit as follows $\displaystyle \lim_{x \rightarrow 0} k \cdot \frac{\sin(kx)}{kx}$. Now, it seems obvious that we can define $z = kx$ and rewrite the limit as $\displaystyle k \cdot \lim_{z \rightarrow 0} \frac{\sin(z)}{z}$ and conclude the limit is equal to $k$. This is a common technique called a change of variables. In this post, we will show when we are allowed to do this.
Limit Law Proof
Let $f: B \rightarrow C$ be a continuous function. This means that $\lim_{z \rightarrow b} f(z) = f(b)$ for all $b \in B$.
Let $g: A \rightarrow B$ be a function (not necessarily continuous).
Let $b = \lim_{x \rightarrow a} g(x)$.
Therefore,
\[\lim_{x \rightarrow a} f(g(x)) = f \left ( \lim_{x \rightarrow a} g(x) \right ) = f(b) = \lim_{z \rightarrow b} f(z)\]Counter-Example
Some sources will say that you can use change of variables as long as $\displaystyle \lim_{z \rightarrow b} f(z)$ exists. However, this is not true. I show a counter-example below.
Suppose \(f(z) = \begin{cases} 1 &\quad\text{if } z = 0 \\ 0 &\quad\text{otherwise} \end{cases}\) and \(g(x) = x \sin(1/x)\)
Clearly, \(\displaystyle \lim_{z \rightarrow 0} f(z) = 0\). I will leave it as an exercise to the reader to show that \(\displaystyle \lim_{x \rightarrow 0} g(x) = 0\).
Now, consider \(f(g(x)) = \begin{cases} 1 &\quad\text{if } x = \frac{1}{n\pi} \text{ for some } n \in \mathbb{Z} \\ 0 &\quad\text{otherwise} \end{cases}\).
Therefore, \(\displaystyle \lim_{x \rightarrow 0} f(g(x))\) does not exist, because the function oscillates infinitely between $0$ and $1$ at $x = 0$.
Comment on the Motivating Example
You may be wondering, how could we know that $\displaystyle \lim_{x \rightarrow 0} \frac{\sin(kx)}{x}$ was continuous before applying this change of variable law? Doesn’t this just move the problem back one step? Actually, we can know that it’s continuous. In a future post, we show that $\sin$ is continuous. We know that the identity function is continuous. Finally, in another future post we show that division preserves continuity. Thus, we do know that this function is continuous and can apply change of variables without circularity.