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 \[ \lim_{x\to \frac{\pi}{2}}\frac{\sin\left( \cos\left( x \right) \right)}{\cos(x)} = \lim_{y\to 0} \frac{\sin x}{x} = 1 \]
\(\displaystyle \lim_{x\to -2} \frac{-3 x - 6}{\sqrt[3]{x + 10 } - 2}={}\)\(-36\)
This is a limit of the indeterminate form \(\frac{0}{0}\). A change of variable \(y = \sqrt[3]{ x + 10 } \) gives \(y^3 = x + 10\), i.e. \(x = y^3 - 10\). The function becomes: \[ \frac{-3 x - 6}{\sqrt[3]{x + 10 } - 2} = \frac{-3 y^3 + 24}{y-2}\] We see that \(x\to-2\) corresponds to \(y\to \sqrt[3]{ -2 + 10 } = \sqrt[3]{8} = 2\). Hence the requested limit equals: \[\begin{aligned}\lim_{x\to -2} \frac{-3 x - 6}{\sqrt[3]{x + 10 } - 2} &= \lim_{y\to 2} \frac{-3 y^3 + 24}{y-2} \\[0.25cm]&= -3 \cdot\lim_{y\to 2} \frac{y^3-8}{y-2} \\[0.25cm] &=-3 \cdot\lim_{y\to 2} \frac{\left(y-2\right) \left( y^2 +2 y +4 \right)}{y-2} \\[0.25cm] &=-3 \cdot\lim_{y\to 2} \left( y^2 +2 y +4 \right) \\[0.25cm] &= -36\end{aligned}\]
In the next paragraph we will cover the use of substitutions for one-sided limits.
