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 \[ \lim_{x\to \frac{\pi}{2}}\frac{\sin\left( \cos\left( x \right) \right)}{\cos(x)} = \lim_{y\to 0} \frac{\sin x}{x} = 1 \]
\(\displaystyle \lim_{x\to -3} \frac{-x - 3}{\sqrt[3]{x + 67 } - 4}={}\)\(-48\)
This is a limit of the indeterminate form \(\frac{0}{0}\). A change of variable \(y = \sqrt[3]{ x + 67 } \) gives \(y^3 = x + 67\), i.e. \(x = y^3 - 67\). The function becomes: \[ \frac{-x - 3}{\sqrt[3]{x + 67 } - 4} = \frac{-y^3 + 64}{y-4}\] We see that \(x\to-3\) corresponds to \(y\to \sqrt[3]{ -3 + 67 } = \sqrt[3]{64} = 4\). Hence the requested limit equals: \[\begin{aligned}\lim_{x\to -3} \frac{-x - 3}{\sqrt[3]{x + 67 } - 4} &= \lim_{y\to 4} \frac{-y^3 + 64}{y-4} \\[0.25cm]&= -1 \cdot\lim_{y\to 4} \frac{y^3-64}{y-4} \\[0.25cm] &=-1 \cdot\lim_{y\to 4} \frac{\left(y-4\right) \left( y^2 +4 y +16 \right)}{y-4} \\[0.25cm] &=-1 \cdot\lim_{y\to 4} \left( y^2 +4 y +16 \right) \\[0.25cm] &= -48\end{aligned}\]
In the next paragraph we will cover the use of substitutions for one-sided limits.
