L'Hôpital's Rule

L’Hôpital’s rule is used to evaluate Limit of the indeterminate form $\frac{\infty }{\infty }$, as long as both the numerator and denominator functions are differentiable. It’s more tedious, so check if common terms can be factored out or cancelled.

Definition §

Let $f,g$ be real-valued functions.

$$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$$

• The function must be in the form $\frac{\infty }{\infty }$ or $\frac{0}{0}$ .
• Given an open interval ((3,4), (a,b), etc.) $D$ where $x\ne a$:
  • $f$ and $g$ must be differentiable in $D$
  • $g^{\prime }(x)\ne 0$ in $D$
• Obviously, ${\lim }_{x\to a}\frac{f^{\prime }(x)}{g^{\prime }(x)}$ needs to exist

The rule also works for one-sided Limits.

Uses In Proving Limits §

L’Hôpital’s rule can be used to prove a limit exists, but it cannot be used to show that a limit does not exist. I.e. there are functions which the rule does not work for, but the limit does exist:

Counterexample

Evaluate the following limit: ${\lim }_{x\to \infty }(\frac{x+\sin (x)}{x})$
First, let’s try using L’hôpital’s rule, since we know that $f(x)=x+\sin (x)$ and $g(x)=x$ are both differentiable and the limit is of the indeterminate form $\frac{\infty }{\infty }$: \

As $\cos (x)$ oscillates between -1 and 1 as x approaches infinity, the limit oscillates between 0 and 2. However, since a limit must be a single, unique value, L’hôpital’s rule says that the limit doesn’t exit.
Or does it? Let’s try using the Sandwich Theorem:
${\lim }_{x\to \infty }(\frac{x+\sin (x)}{x})={\lim }_{x\to \infty }(1+\frac{\sin (x)}{x})$ $=1+{\lim }_{x\to \infty }\left(\frac{\sin (x)}{x}\right)$ (using Limit Laws) $\sin (x)\in [-1,1]\forall x$ $⟹\frac{\sin (x)}{x}\in [-\frac{1}{x},\frac{1}{x}]$ ${\lim }_{x\to \infty }\left(\frac{1}{x}\right)={\lim }_{x\to \infty }\left(-\frac{1}{x}\right)=0$ (by Standard Limits) Hence, the sandwich theorem applies, since $-\frac{1}{x}\le \frac{\sin (x)}{x}\le \frac{1}{x}$ $1+{\lim }_{x\to \infty }\left(\frac{\sin (x)}{x}\right)=1+0=1$
${\lim }_{x\to \infty }\left(\frac{x+\sin (x)}{x}\right)=1$