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 \[ \lim_{x\to \frac{\pi}{2}}\frac{\sin\left( \cos\left( x \right) \right)}{\cos(x)} = \lim_{y\to 0} \frac{\sin x}{x} = 1 \]
\(\displaystyle \lim_{x\to -4} \frac{2 x +8}{\sqrt[3]{x -60 } +4}={}\)\(96\)
This is a limit of the indeterminate form \(\frac{0}{0}\). A change of variable \(y = \sqrt[3]{ x -60 } \) gives \(y^3 = x -60\), i.e. \(x = y^3 +60\). The function becomes: \[ \frac{2 x +8}{\sqrt[3]{x -60 } +4} = \frac{2 y^3 + 128}{y+4}\] We see that \(x\to-4\) corresponds to \(y\to \sqrt[3]{ -4 -60 } = \sqrt[3]{-64} = -4\). Hence the requested limit equals: \[\begin{aligned}\lim_{x\to -4} \frac{2 x +8}{\sqrt[3]{x -60 } +4} &= \lim_{y\to -4} \frac{2 y^3 + 128}{y+4} \\[0.25cm]&= 2 \cdot\lim_{y\to -4} \frac{y^3+64}{y+4} \\[0.25cm] &=2 \cdot\lim_{y\to -4} \frac{\left(y+4\right) \left( y^2 -4 y +16 \right)}{y+4} \\[0.25cm] &=2 \cdot\lim_{y\to -4} \left( y^2 -4 y +16 \right) \\[0.25cm] &= 96\end{aligned}\]
In the next paragraph we will cover the use of substitutions for one-sided limits.
