Evaluating limits using change of variables

Say, we want to calculate the limit of $f(x) = \frac{2x}{\sqrt[3]{x+27} - 3}$ as $x\to 0$ .

We make use of the following change of variables technique.

Say, $t(x) = \sqrt[3]{x+27}$ . Observe that $t(x)\to 3$ as $x\to 0$ and $f(x)= \frac{2(t^3(x) - 27)}{t(x) - 3}$ so that we write:

$\lim\limits_{x\to 0} \frac{2x}{\sqrt[3]{x+27} - 3} = \lim\limits_{t\to 3}\frac{2(t^3 - 27)}{t - 3} = \lim\limits_{t\to 3} 2(t^2+3t+9) = 54$

The crucial step is being able to change $\lim\limits_{x\to 0} \frac{2x}{\sqrt[3]{x+27} - 3}$ to $\lim\limits_{t\to 3}\frac{2(t^3 - 27)}{t - 3}$ .

What is the justification behind making this change? Say, we want to calculate $\lim\limits_{x\to a}f[t(x)]$ knowing that $\lim\limits_{x\to a} t(x) = t_a$ and that $\lim\limits_{t\to t_a} f(t)$ exists. Can we write $\lim\limits_{x\to a}f[t(x)] = \lim\limits_{t\to t_a} f(t)$ ? NOT ALWAYS!

For a counter example, say,

$f(x) = \begin{cases} 2, & \mbox{if }x\neq 1 \\ 1, & \mbox{if }x = 1\end{cases}$

Observe that $\lim\limits_{x\to 1}f(x) = 2$ .

Now, define $t(x) = 1$ so that $\lim\limits_{x\to 0} t(x) = 1$ .

Now, observe that $f[t(x)] = 1$ so that $\lim\limits_{x\to 0} f[t(x)] = 1$ . But, using the change of variables technique, $\lim_{t\to 1}f(t)$ yields $2$ .

Theorem: Let $t:\bf{R}\to\bf{R}$ and $f:\bf{R}\to\bf{R}$ be two functions. Say, $\lim\limits_{x\to a} t(x) = t_a$ , but $t(x)$ does not take the value $t_a$ around $a$ . (That is, there exists an open interval around $a$ namely $(a-\alpha, a+\alpha)$ such that $\forall x\in(a-\alpha,a+\alpha)\setminus\{a\}, t(x)\neq t_a)$ . Further, say, $\lim\limits_{t\to t_a} f(t)$ exists. Then,

$\lim\limits_{x\to a}f[t(x)] = \lim\limits_{t\to t_a}f(t)$ .

Proof:

Say, $\{x_n\}$ is a sequence in $\bf{R}\setminus\{a\}$ such that $\{x_n\}\to a$ . Choose $N\in\bf{N}$ such that $\forall n\ge N$ , $x_n\in(a-\alpha,a+\alpha)$ .

Since $\lim\limits_{x\to a} t(x) = t_a$ , the sequence $\{t(x_n)\}\to t_a$ and $\forall n\ge N, t(x_n)\ne t_a$ .

Now, take the tail $\{t(x_{N+k})\} = \{t(x_{N+1}), t(x_{N+2}),\ldots\}$ such that $\{t(x_{N+k})\}\to t_a$ and $\forall k, t(x_{N+k})\ne t_a$ .

Then, by the definition of the limit, $\{f[t(x_{N+k})]\}\to\lim\limits_{t\to t_a}f(t)$ . And hence, $\{f[t(x_{n})]\}\to\lim\limits_{t\to t_a}f(t)$ .

Thus, for any sequence $\{x_n\}$ in $\bf{R}\setminus\{a\}$ such that $\{x_n\}\to a$ , $\{f[t(x_{n})]\}\to\lim\limits_{t\to t_a}f(t)$ .

Thus, $\lim\limits_{x\to a}f[t(x)] = \lim \limits_{t\to t_a}f(t)$ .

Note: It is crucial that $t(x)$ does not take the value $t_a$ around $a$ . This is because $f(t_a)$ may not be equal to $\lim\limits_{t\to t_a}f(t)$ . This restriction is not an obstacle in practice because all we need is a small interval around $a$ wherein for all values besides $a$ , $t$ takes values different from $t_a$ . If this restriction is violated by our substitution $t(x)$ , this implies $t$ takes the value $t_a$ for infinite number of values around $a$ . This does not happen for usual substitutions.

Note: If $f(t)$ is continuous at $t_a$ , $f(t_a)$ is defined and $\lim\limits_{t\to t_a}f(t) = f(t_a)$ so that in this case we do not need the restriction that $t(x)$ not take the value $t_a$ . In this case, $\lim\limits_{x\to a}f[t(x)] = f(t_a)$ . Still, it seems to me, the value of change of variables technique only when $f(t)$ is not continuous at $t_a$ . This is because if our substitution $t(x)$ is also continuous at $a$ (which I think is typically the case), $f[t(x)]$ is continuous by theorem on composition of continuous functions. In this case, we could just evaluate $\lim_{x\to a}f[t(x)]$ merely as $f[t(a)]$ . There is no need to use change of variables or any special limit evaluation method.

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