L'Hôpital's rule

We have been using the following formula without proof so far:

$$\underset{x\to 0}{lim}\frac{\sin x}{x}=1.$$

In this example, both $\sin x$ and $x$ converges to 0 as $x\to 0$ so we have something like $\frac{0}{0}$. In general, if $\underset{x\to a}{lim}f(x)=\underset{x\to a}{lim}g(x)=0$ or $\underset{x\to a}{lim}f(x)=\underset{x\to a}{lim}g(x)=\pm \mathrm{\infty }$, the limit of the form $\underset{x\to a}{lim}\frac{f(x)}{g(x)}$ is called an indeterminate form. L'Hôpital's rule provides a convenient way to calculate such limits (The proof will be given in another post).

See also: Proof of L'Hôpital's rule

Theorem (L'Hôpital's rule (1))

Let $f(x)$ and $g(x)$ be differentiable functions on the open interval $(a,b)$ that satisfy the following conditions.

1. $$\underset{x\to a+0}{lim}f(x)=\underset{x\to a+0}{lim}g(x)=0.$$
2. For all $x\in (a,b)$, $g^{\prime }(x)\ne 0$.
3. The right limit $$\underset{x\to a+0}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$$ exists.

Then, the right limit $\underset{x\to a+0}{lim}\frac{f(x)}{g(x)}$ exists and

$$\underset{x\to a+0}{lim}\frac{f(x)}{g(x)}=\underset{x\to a+0}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}.$$

Furthermore, condition 1 may be replaced with

• (1') $$\underset{x\to a+0}{lim}f(x)=\pm \mathrm{\infty }\text{ and }\underset{x\to a+0}{lim}g(x)=\pm \mathrm{\infty }$$

to have the same conclusion.

Also the right limits $\underset{x\to a+0}{lim}$ may be replaced with the left limits $\underset{x\to a-0}{lim}$.

Corollary (L'Hôpital's rule (2))

Let $f(x)$ and $g(x)$ be differentiable functions on an open interval $I$ that contains $a$. Suppose $f(x)$ and $g(x)$ satisfy the following conditions.

1. $$\underset{x\to a}{lim}f(x)=\underset{x\to a}{lim}g(x)=0.$$
2. For all $x\in I\setminus \{a\}$, $g^{\prime }(x)\ne 0$.
3. The limit $\underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$ exists.

Then, the limit $\underset{x\to a}{lim}\frac{f(x)}{g(x)}$ exists and

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

Furthermore, condition 1 may be replaced with

• (1') $$\underset{x\to a}{lim}f(x)=\pm \mathrm{\infty }\text{ and }\underset{x\to a}{lim}g(x)=\pm \mathrm{\infty }$$

to have the same conclusion.

Example. Now let us prove

$$\underset{x\to 0}{lim}\frac{\sin x}{x}=1$$

by applying L'Hôpital's rule. First, we need to check the conditions.

Both $\sin x$ and $x$ are defined on $\mathbb{R}$ which is an open interval.

1. We have $$\underset{x\to 0}{lim}\sin x=0$$ and $$\underset{x\to 0}{lim}x=0.$$
2. $(x{)}^{\prime }=1\ne 0$ for all $x\in \mathbb{R}$.
3. We know $(\sin x{)}^{\prime }=\cos x$ and $\underset{x\to 0}{lim}\cos x=1$ so $$\underset{x\to 0}{lim}\frac{(\sin x{)}^{\prime }}{(x{)}^{\prime }}=\underset{x\to 0}{lim}\frac{\cos x}{1}=1.$$

Therefore, all the conditions are satisfied and the limit $\underset{x\to 0}{lim}\frac{\sin x}{x}$ exists and

$$\underset{x\to 0}{lim}\frac{\sin x}{x}=\underset{x\to 0}{lim}\frac{(\sin x{)}^{\prime }}{(x{)}^{\prime }}=\underset{x\to 0}{lim}\frac{\cos x}{1}=1.$$

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Example. Consider

$$\underset{x\to 0}{lim}\frac{\log (\cos x)}{x^2}.$$

$\underset{x\to 0}{lim}\log (\cos x)=0$ and $\underset{x\to 0}{lim}{x}^2=0$ and if $x\ne 0$ then $(x^2{)}^{\prime }=2x\ne 0$. Furthermore,

$$\begin{array}{rcl}\underset{x\to 0}{lim}\frac{\{\log (\cos x){\}}^{\prime }}{\{x^2{\}}^{\prime }}& =& \underset{x\to 0}{lim}(\frac{-\sin x}{\cos x}\cdot \frac{1}{2x})\\ & =& \underset{x\to 0}{lim}(\frac{\sin x}{x}\cdot \frac{-1}{2\cos x})\\ & =& -\frac{1}{2}.\end{array}$$

Therefore,

$$\underset{x\to 0}{lim}\frac{\log (\cos x)}{x^2}=-\frac{1}{2}.$$

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Corollary (L'Hôpital's rule (3))

Let $f(x)$ and $g(x)$ be differentiable functions on the open interval $(b,\mathrm{\infty })$ that satisfy the following conditions.

1. $$\underset{x\to \mathrm{\infty }}{lim}f(x)=\underset{x\to \mathrm{\infty }}{lim}g(x)=0.$$
2. For all $x>b$, $g^{\prime }(x)\ne 0$.
3. The limit $\underset{x\to \mathrm{\infty }}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$ exists.

Then, the limit $\underset{x\to \mathrm{\infty }}{lim}\frac{f(x)}{g(x)}$ also exists and

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

Furthermore, condition 1 may be replaced with

• (1') $$\underset{x\to \mathrm{\infty }}{lim}f(x)=\pm \mathrm{\infty }\text{ and }\underset{x\to \mathrm{\infty }}{lim}g(x)=\pm \mathrm{\infty }$$

to have the same conclusion.

Remark. The same corollary holds if $f(x)$ and $g(x)$ are continuous functions on $(-\mathrm{\infty },b)$ and the limits $\underset{x\to \mathrm{\infty }}{lim}$ are replaced with $\underset{x\to -\mathrm{\infty }}{lim}$. □

Example. Let us find $\underset{x\to \mathrm{\infty }}{lim}{x}^{\frac{1}{x}}$.

Let $f(x)=x^{\frac{1}{x}}$ so that $\log f(x)=\frac{\log x}{x}$. Now consider $\underset{x\to \mathrm{\infty }}{lim}\log f(x)$.

1. $\underset{x\to \mathrm{\infty }}{lim}\log x=\mathrm{\infty }$ and $\underset{x\to \mathrm{\infty }}{lim}x=\mathrm{\infty }$.
2. $(x{)}^{\prime }=1\ne 0$.
3. $$\underset{x\to \mathrm{\infty }}{lim}\frac{(\log x{)}^{\prime }}{(x{)}^{\prime }}=\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x}=0.$$

Thus,

$$\underset{x\to \mathrm{\infty }}{lim}\frac{\log x}{x}=0.$$

Noting that $f(x)=e^{\log f(x)}$ and that the exponential function is continuous, we find

$$\underset{x\to \mathrm{\infty }}{lim}{x}^{\frac{1}{x}}=\underset{x\to \mathrm{\infty }}{lim}{e}^{\frac{\log x}{x}}=e^0=1.$$

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