Proof of L'Hôpital's rule

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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 \(g'(x) \neq 0\) for all \(x \in (a,b)\).  Then, there exists a \(c\in (a,b)\) such that

\[\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}.\]

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

\[g'(d) = \frac{g(b) - g(a)}{b - a}.\]

Since \(g'(d) \neq 0\) by assumption, it follows that \(g(b) - g(a) \neq 0\). 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 \(h'(c) = 0\). Then

\[h'(c) = f'(c) - \frac{f(b) - f(a)}{g(b) - g(a)}g'(c) = 0\]

so that, since \(g'(c) \neq 0\),

\[\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}.\]

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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. \[\lim_{x\to a+0}f(x) = \lim_{x\to a+0}g(x) = 0.\]
  2. For all \(x \in (a,b)\), \(g'(x) \neq 0\).
  3. The right limit \[\lim_{x\to a+0}\frac{f'(x)}{g'(x)}\] exists.
Then, the right limit \(\lim_{x\to a+0}\frac{f(x)}{g(x)}\) exists and
\[\lim_{x\to a+0}\frac{f(x)}{g(x)} = \lim_{x\to a+0}\frac{f'(x)}{g'(x)}.\]
Furthermore, Condition 1 may be replaced with
\[ \lim_{x\to a+0}f(x) = \lim_{x\to a+0}g(x) = \pm \infty. \]
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 \(t \in (a,x)\), \(g'(t) \neq 0\). Thus, by Cauchy's mean value theorem, there exists a \(c_x \in (a,x)\) such that
\[\frac{f'(c_x)}{g'(c_x)} = \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f(x)}{g(x)}.\]
\(c_x \to a + 0\) as \(x \to a +0\) and, by condition 3, \(\lim_{x\to a+0}\frac{f'(c_x)}{g'(c_x)}\) exists. Therefore \(\lim_{x\to a+0}\frac{f(x)}{g(x)}\) also exists and
\[\lim_{x\to a+0}\frac{f(x)}{g(x)} = \lim_{x\to a+0}\frac{f'(c_x)}{g'(c_x)} =  \lim_{x\to a+0}\frac{f'(x)}{g'(x)}.\]

Next, consider the case when
  • (1') \[\lim_{x\to a+0}f(x) = \lim_{x\to a+0}g(x) = \pm \infty\]
holds instead of condition 1 above.
Let us define the constant \(L\) by
\[L = \lim_{x \to a+0}\frac{f'(x)}{g'(x)}.\]
By the definition of the right limit, for any \(\varepsilon > 0\), there exists a \(\delta_1 > 0\) such that \(a < x < a + \delta_1\) implies \(\left|\frac{f'(x)}{g'(x)} - L\right| < \varepsilon\). 
Since \(\lim_{x\to a+0}g(x) = \pm\infty\), there exists \(\delta_2 > 0\) such that \(a < x < a + \delta_2\) implies \(|g(x)| > 1\).
Let \(\delta' = \min\{\delta_1, \delta_2\}\) and \(d = a + \delta'\). By Cauchy's mean value theorem, for all \(x \in (a, d)\),  there exists a \(c_x \in (x, d)\) such that
\[\frac{f'(c_x)}{g'(c_x)} = \frac{f(d) - f(x)}{g(d) - g(x)} = \frac{\frac{f(x)}{g(x)} - \frac{f(d)}{g(x)}}{1 - \frac{g(d)}{g(x)}}.\]
By rearranging this equation, we have
\[\frac{f'(c_x)}{g'(c_x)} = \frac{f(x)}{g(x)} - r(x)\tag{eq:rx}\]
where
\[r(x) = \frac{f(d)}{g(x)} - \frac{f'(c_x)}{g'(c_x)}\cdot\frac{g(d)}{g(x)}.\]
Here, \(f(d)\) and \(g(d)\) are finite constants, and \(\lim_{x\to a+0}\frac{f'(x)}{g'(x)}\) converges to a finite value (by condition 3). Hence, by condition 1' (\(\lim_{x\to a + 0}g(x) = \pm\infty\)), \(\lim_{x\to a+0}r(x) = 0\). In other words, for any \(\varepsilon > 0\), there exists a \(\delta_3 > 0\) such that \(a < x < a + \delta_3\) implies \(|r(x)| < \varepsilon\).
Let \(\delta = \min\{\delta', \delta_3\}\). By Eq. (eq:rx), we have
\[\frac{f(x)}{g(x)} - L = \frac{f'(c_x)}{g'(c_x)} - L + r(x).\]
\(a < x < a + \delta\) implies 
\[ \begin{equation} \left|\frac{f(x)}{g(x)} - L\right| \leq \left|\frac{f'(c_x)}{g'(c_x)} - L\right| + |r(x)| < 2\varepsilon. \end{equation} \]
Therefore the right limit \(\lim_{x\to a+0}\frac{f(x)}{g(x)}\) exists and it is equal to \(L\). ■
Remark. The case of left limits \(\lim_{x\to a-0}\) 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 \((b, \infty)\) that satisfy the following conditions.
  1. \[\lim_{x \to\infty}f(x) = \lim_{x\to\infty}g(x) = 0.\]
  2. For all \(x > b\), \(g'(x) \neq 0\).
  3. The limit \(\lim_{x\to \infty}\frac{f'(x)}{g'(x)}\) exists.
Then, the limit \(\lim_{x\to \infty}\frac{f(x)}{g(x)}\) also exists and
\[\lim_{x\to \infty}\frac{f(x)}{g(x)} = \lim_{x\to \infty}\frac{f'(x)}{g'(x)}.\]
Furthermore, condition 1 may be replaced with
  • (1') \[\lim_{x\to \infty}f(x) = \pm\infty \text{ and } \lim_{x\to \infty}g(x) =\pm\infty\]
to have the same conclusion.

Proof. We prove the case when \(x \to \infty\) and condition 1 holds.
For the open interval \((b, \infty)\), 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 \(x \to \infty\), \(t\to +0\). By conditions 1 and 2,
\[\lim_{t\to +0}f\left(\frac{1}{t}\right) = \lim_{t\to +0}g\left(\frac{1}{t}\right) = 0\]
and
\[g'\left(\frac{1}{t}\right) \neq 0 ~ \text{ for all } t \in \left(0, \frac{1}{b}\right).\]
From
\[\frac{d}{dt}f\left(\frac{1}{t}\right) = -\frac{1}{t^2}f'\left(\frac{1}{t}\right)\]
and
\[\frac{d}{dt}g\left(\frac{1}{t}\right) = -\frac{1}{t^2}g'\left(\frac{1}{t}\right),\]
we have
\[\lim_{t\to +0}\frac{\frac{d}{dt}f\left(\frac{1}{t}\right)}{\frac{d}{dt}g\left(\frac{1}{t}\right)} = \lim_{t \to +0}\frac{f'\left(\frac{1}{t}\right)}{g'\left(\frac{1}{t}\right)} = \lim_{x\to \infty}\frac{f'(x)}{g'(x)}\]
so that, by condition 3, the right limit \(\lim_{t\to +0}\frac{\frac{d}{dt}f\left(\frac{1}{t}\right)}{\frac{d}{dt}g\left(\frac{1}{t}\right)}\) exists. Therefore, by L'H\^opital's rule (1), we have the limit
\[\lim_{t\to +0}\frac{f\left(\frac{1}{t}\right)}{g\left(\frac{1}{t}\right)} = \lim_{x\to\infty}\frac{f(x)}{g(x)} = \lim_{x\to \infty}\frac{f'(x)}{g'(x)}.\]
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