Substitution rule

Limits part 2: Functions: Techniques

Substitution rule

Recall that for continuous functions $f$ we have

$\lim_{y\to a} f\left(g(y)\right) = f\left( \lim_{y\to a} g(y)\right)$ If we apply this twice we obtain

$\lim_{y\to a} f\left(g(y)\right) = f\left( \lim_{y\to a} g(y) \right) = \lim_{x \to \lim_{y\to a} g(y)} f(x)$

Substitutions are useful to calculate limits of compositions. Given a limit $\lim_{x\to b} f(x)$ we can select a function $g$ and a point $a$ satisfying $\lim_{y\to a} g(y) = b$ . Next we get:

$\lim_{x \to b} f(x) = \lim_{y\to a} f\left(g(y)\right)$ Substitution can also be used for limits to $\pm\infty$ .

In the next two examples we use substitutions to evaluate limits:

It holds that

$\lim_{x\to \frac{\pi}{2}}\frac{\sin\left( \cos\left( x \right) \right)}{\cos(x)} = \lim_{y\to 0} \frac{\sin x}{x} = 1$

In this case we have $f(x) = \frac{\sin(x)}{x}$ and $g(y) = \cos(y)$ .
Furthermore we have $a = \frac{\pi}{2}$ and $b = \lim_{y\to a} g(y) = \lim_{y\to\frac{\pi}{2}}\cos(y) = 0$ .

Calculate

$\lim_{x\to -1} \frac{3 x +3}{\sqrt[3]{x + 65 } - 4}$

$\displaystyle \lim_{x\to -1} \frac{3 x +3}{\sqrt[3]{x + 65 } - 4}={}$ $144$

This is a limit of the indeterminate form $\frac{0}{0}$ . A change of variable $y = \sqrt[3]{ x + 65 }$ gives $y^3 = x + 65$ , i.e. $x = y^3 - 65$ . The function becomes:

$\frac{3 x +3}{\sqrt[3]{x + 65 } - 4} = \frac{3 y^3 -192}{y-4}$ We see that $x\to-1$ corresponds to $y\to \sqrt[3]{ -1 + 65 } = \sqrt[3]{64} = 4$ . Hence the requested limit equals:

$\begin{aligned}\lim_{x\to -1} \frac{3 x +3}{\sqrt[3]{x + 65 } - 4} &= \lim_{y\to 4} \frac{3 y^3 -192}{y-4} \\[0.25cm]&= 3 \cdot\lim_{y\to 4} \frac{y^3-64}{y-4} \\[0.25cm] &=3 \cdot\lim_{y\to 4} \frac{\left(y-4\right) \left( y^2 +4 y +16 \right)}{y-4} \\[0.25cm] &=3 \cdot\lim_{y\to 4} \left( y^2 +4 y +16 \right) \\[0.25cm] &= 144\end{aligned}$

New example

In the next paragraph we will cover the use of substitutions for one-sided limits.