Substitution and one-sided limits

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 $-\infty$ .