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 \[ \lim_{x\to \frac{\pi}{2}}\frac{\sin\left( \cos\left( x \right) \right)}{\cos(x)} = \lim_{y\to 0} \frac{\sin x}{x} = 1 \]
\(\displaystyle \lim_{x\to -4} \frac{-2 x - 8}{\sqrt[3]{x + 5 } - 1}={}\)\(-6\)
This is a limit of the indeterminate form \(\frac{0}{0}\). A change of variable \(y = \sqrt[3]{ x + 5 } \) gives \(y^3 = x + 5\), i.e. \(x = y^3 - 5\). The function becomes: \[ \frac{-2 x - 8}{\sqrt[3]{x + 5 } - 1} = \frac{-2 y^3 + 2}{y-1}\] We see that \(x\to-4\) corresponds to \(y\to \sqrt[3]{ -4 + 5 } = \sqrt[3]{1} = 1\). Hence the requested limit equals: \[\begin{aligned}\lim_{x\to -4} \frac{-2 x - 8}{\sqrt[3]{x + 5 } - 1} &= \lim_{y\to 1} \frac{-2 y^3 + 2}{y-1} \\[0.25cm]&= -2 \cdot\lim_{y\to 1} \frac{y^3-1}{y-1} \\[0.25cm] &=-2 \cdot\lim_{y\to 1} \frac{\left(y-1\right) \left( y^2 +y +1 \right)}{y-1} \\[0.25cm] &=-2 \cdot\lim_{y\to 1} \left( y^2 +y +1 \right) \\[0.25cm] &= -6\end{aligned}\]
In the next paragraph we will cover the use of substitutions for one-sided limits.
