### Limits part 2: Functions: Techniques

### Substitution and one-sided limits

We just saw how we can use substitutions to calculate limits. In this paragraph we will consider one-sided limits.

Suppose that for some function \(f\) we would like to calculate the value of \[ \lim_{x\to 0} f(x)\] It makes sense to substitute \(y = \frac{1}{x}\) and determine the limit for \(y\to\infty\). A problem we encounter now is that \(\frac{1}{x}\) approaches \(0\) only from above so this notion belongs to \(\lim_{x\downarrow 0}\). This means the limit unintentionally changes into a one-sided limit:\[ \lim_{x\downarrow 0} f(x) = \lim_{y\to\infty} f\left(\dfrac{1}{y}\right)\] To ensure \(\lim_{x\to 0}f(x)\) exists we should also determine \(\lim_{x\uparrow 0} f(x) \). An option is to substitute \( z = -\frac{1}{x}\) in order to determine the left limit.

Look at \[ \lim_{x\downarrow 0} \sqrt[x]{1+x} = \lim_{x\downarrow 0}\left( 1+x \right)^\frac{1}{x}\]

When we substitute \(y = \frac{1}{x}\) this changes into \[ \lim_{y\to\infty} \left( 1 + \frac{1}{y}\right)^y\] This is a standard limit with value \(e\).

In this example we have \[f(x) = \sqrt[x]{1+x}, \qquad g(y) = \frac{1}{y}, \qquad a = \infty, \qquad b = 0\] in the substitution rule. The reason that \(b = 0\) is that \(\frac{1}{y}\) approaches zero from above when \(y\) approaches \(\infty\). When you substitute \(\frac{1}{y}\) and you want to say something about the limit at zero, then you must explore the behaviour at both \(\infty\) and \(-\infty\).