Proof of L'Hôpital's rule

To prove L'Hôpital's rule, we first prove Cauchy's mean value theorem that generalizes the mean value theorem provided earlier.

See also: Mean Value Theorem

Theorem (Cauchy's Mean Value Theorem)

Let $f(x)$ and $g(x)$ be functions that are continuous on $[a,b]$ and differentiable on $(a,b)$. Suppose that $g^{\prime }(x)\ne 0$ for all $x\in (a,b)$.  Then, there exists a $c\in (a,b)$ such that

$$\frac{f^{\prime }(c)}{g^{\prime }(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}.$$

Proof. Note that the function $g(x)$ satisfies the conditions of the mean value theorem. Thus, there exists a $d\in (a,b)$ such that

$$g^{\prime }(d)=\frac{g(b)-g(a)}{b-a}.$$

Since $g^{\prime }(d)\ne 0$ by assumption, it follows that $g(b)-g(a)\ne 0$. Now let us define

$$h(x)=f(x)-\frac{f(b)-f(a)}{g(b)-g(a)}g(x).$$

Then $h(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Moreover,

$$h(a)=\frac{f(a)g(b)-f(b)g(a)}{g(b)-g(a)}=h(b).$$

Therefore, by Rolle's theorem (Theorem 10.6), there exists a $c\in (a,b)$ such that $h^{\prime }(c)=0$. Then

$$h^{\prime }(c)=f^{\prime }(c)-\frac{f(b)-f(a)}{g(b)-g(a)}{g}^{\prime }(c)=0$$

so that, since $g^{\prime }(c)\ne 0$,

$$\frac{f^{\prime }(c)}{g^{\prime }(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}.$$

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Theorem L'Hôpital's rule (1)

Let $f(x)$ and $g(x)$ be differentiable functions on $(a,b)$. Assume that the following conditions are satisfied.

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

$$\underset{x\to a+0}{lim}f(x)=\underset{x\to a+0}{lim}g(x)=\pm \mathrm{\infty }.$$

Proof. For convenience we set $f(a)=g(a)=0$ so that $f(x)$ and $g(x)$ are defined on $[a,b)$. Assume conditions 1, 2, and 3 hold.

By condition 1, $f(x)$ and $g(x)$ are continuous on $[a,b)$. For any $x\in (a,b)$, $f(x)$ and $g(x)$ are continuous on $[a,x]$ and differentiable on $(a,x)$. By condition 2, for all $t\in (a,x)$, $g^{\prime }(t)\ne 0$. Thus, by Cauchy's mean value theorem, there exists a $c_x\in (a,x)$ such that

$$\frac{f^{\prime }(c_x)}{g^{\prime }(c_x)}=\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f(x)}{g(x)}.$$

$c_x\to a+0$ as $x\to a+0$ and, by condition 3, $\underset{x\to a+0}{lim}\frac{f^{\prime }(c_x)}{g^{\prime }(c_x)}$ exists. Therefore $\underset{x\to a+0}{lim}\frac{f(x)}{g(x)}$ also exists and

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

Next, consider the case when

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

holds instead of condition 1 above.

Let us define the constant $L$ by

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

By the definition of the right limit, for any $\epsilon >0$, there exists a ${\delta }_1>0$ such that $a<x<a+{\delta }_1$ implies $|\frac{f^{\prime }(x)}{g^{\prime }(x)}-L|<\epsilon$. 

Since $\underset{x\to a+0}{lim}g(x)=\pm \mathrm{\infty }$, there exists ${\delta }_2>0$ such that $a<x<a+{\delta }_2$ implies $|g(x)|>1$.

Let ${\delta }^{\prime }=min\{{\delta }_1,{\delta }_2\}$ and $d=a+{\delta }^{\prime }$. By Cauchy's mean value theorem, for all $x\in (a,d)$,  there exists a $c_x\in (x,d)$ such that

$$\frac{f^{\prime }(c_x)}{g^{\prime }(c_x)}=\frac{f(d)-f(x)}{g(d)-g(x)}=\frac{\frac{f(x)}{g(x)}-\frac{f(d)}{g(x)}}{1-\frac{g(d)}{g(x)}}.$$

By rearranging this equation, we have

$$\begin{array}{}\text{(eq:rx)}& \frac{f^{\prime }(c_x)}{g^{\prime }(c_x)}=\frac{f(x)}{g(x)}-r(x)\end{array}$$

where

$$r(x)=\frac{f(d)}{g(x)}-\frac{f^{\prime }(c_x)}{g^{\prime }(c_x)}\cdot \frac{g(d)}{g(x)}.$$

Here, $f(d)$ and $g(d)$ are finite constants, and $\underset{x\to a+0}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$ converges to a finite value (by condition 3). Hence, by condition 1' ($\underset{x\to a+0}{lim}g(x)=\pm \mathrm{\infty }$), $\underset{x\to a+0}{lim}r(x)=0$. In other words, for any $\epsilon >0$, there exists a ${\delta }_3>0$ such that $a<x<a+{\delta }_3$ implies $|r(x)|<\epsilon$.

Let $\delta =min\{{\delta }^{\prime },{\delta }_3\}$. By Eq. (eq:rx), we have

$$\frac{f(x)}{g(x)}-L=\frac{f^{\prime }(c_x)}{g^{\prime }(c_x)}-L+r(x).$$

$a<x<a+\delta$ implies 

$$|\frac{f(x)}{g(x)}-L|\le |\frac{f^{\prime }(c_x)}{g^{\prime }(c_x)}-L|+|r(x)|<2\epsilon .$$

Therefore the right limit $\underset{x\to a+0}{lim}\frac{f(x)}{g(x)}$ exists and it is equal to $L$. ■

Remark. The case of left limits $\underset{x\to a-0}{lim}$ can be proved similarly. Then, L'Hôpital's rule (2) should be easily proved. (Exercises) □

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.

Proof. We prove the case when $x\to \mathrm{\infty }$ and condition 1 holds.

For the open interval $(b,\mathrm{\infty })$, this $b$ can be replaced with any real number greater than $b$. Therefore, without losing generality, we may assume $b>0$.

Let $x=\frac{1}{t}$. As $x\to \mathrm{\infty }$, $t\to +0$. By conditions 1 and 2,

$$\underset{t\to +0}{lim}f\left(\frac{1}{t}\right)=\underset{t\to +0}{lim}g\left(\frac{1}{t}\right)=0$$

and

$$g^{\prime }\left(\frac{1}{t}\right)\ne 0\text{ for all }t\in (0,\frac{1}{b}).$$

From

$$\frac{d}{dt}f\left(\frac{1}{t}\right)=-\frac{1}{t^2}{f}^{\prime }\left(\frac{1}{t}\right)$$

and

$$\frac{d}{dt}g\left(\frac{1}{t}\right)=-\frac{1}{t^2}{g}^{\prime }\left(\frac{1}{t}\right),$$

we have

$$\underset{t\to +0}{lim}\frac{\frac{d}{dt}f\left(\frac{1}{t}\right)}{\frac{d}{dt}g\left(\frac{1}{t}\right)}=\underset{t\to +0}{lim}\frac{f^{\prime }\left(\frac{1}{t}\right)}{g^{\prime }\left(\frac{1}{t}\right)}=\underset{x\to \mathrm{\infty }}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$$

so that, by condition 3, the right limit $\underset{t\to +0}{lim}\frac{\frac{d}{dt}f\left(\frac{1}{t}\right)}{\frac{d}{dt}g\left(\frac{1}{t}\right)}$ exists. Therefore, by L'H\^opital's rule (1), we have the limit

$$\underset{t\to +0}{lim}\frac{f\left(\frac{1}{t}\right)}{g\left(\frac{1}{t}\right)}=\underset{x\to \mathrm{\infty }}{lim}\frac{f(x)}{g(x)}=\underset{x\to \mathrm{\infty }}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}.$$

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