L'Hôpital's rule

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

\[\lim_{x\to 0}\frac{\sin x}{x} = 1.\]

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 \(\lim_{x\to a}f(x) = \lim_{x\to a}g(x) = 0\) or \(\lim_{x\to a}f(x) = \lim_{x\to a}g(x) = \pm\infty\), the limit of the form \(\lim_{x\to a}\frac{f(x)}{g(x)}\) 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. \[\lim_{x\to a+0}f(x) = \lim_{x\to a+0}g(x) = 0.\]
  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
  • (1') \[\lim_{x\to a+0}f(x) = \pm\infty \text{ and } \lim_{x\to a+0}g(x) = \pm\infty\]
to have the same conclusion.
Also the right limits \(\lim_{x \to a+0}\) may be replaced with the left limits \(\lim_{x\to a-0}\).

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. \[\lim_{x \to a}f(x) = \lim_{x\to a}g(x) = 0.\]
  2. For all \(x \in I\setminus\{a\}\), \(g'(x) \neq 0\).
  3. The limit \(\lim_{x\to a}\frac{f'(x)}{g'(x)}\) exists.
Then, the limit \(\lim_{x\to a}\frac{f(x)}{g(x)}\) exists and
\[\lim_{x\to a}\frac{f(x)}{g(x)} =\lim_{x\to a}\frac{f'(x)}{g'(x)}.\]
Furthermore, condition 1 may be replaced with
  • (1') \[\lim_{x\to a}f(x) = \pm\infty \text{ and } \lim_{x\to a}g(x) =\pm\infty\]
to have the same conclusion.

Example. Now let us prove
\[\lim_{x\to 0}\frac{\sin x}{x}  = 1\]
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 \[\lim_{x\to 0} \sin x = 0\] and \[\lim_{x\to 0}x = 0.\]
  2. \((x)' = 1 \neq 0\) for all \(x\in \mathbb{R}\).
  3. We know \((\sin x)' = \cos x\) and \(\lim_{x\to 0}\cos x = 1\) so \[\lim_{x\to 0}\frac{(\sin x)'}{(x)'} = \lim_{x\to 0}\frac{\cos x}{1} = 1.\]
Therefore, all the conditions are satisfied and the limit \(\lim_{x \to 0}\frac{\sin x}{x}\) exists and
\[\lim_{x \to 0}\frac{\sin x}{x} = \lim_{x\to 0}\frac{(\sin x)'}{(x)'} = \lim_{x\to 0}\frac{\cos x}{1} = 1.\]
□

Example. Consider
\[\lim_{x\to 0}\frac{\log(\cos x)}{x^2}.\]
\(\lim_{x\to 0}\log(\cos x) = 0\) and \(\lim_{x\to 0}x^2 = 0\) and if \(x \neq 0\) then \((x^2)' = 2x \neq 0\). Furthermore,
\[ \begin{eqnarray} \lim_{x\to 0}\frac{\{\log(\cos x)\}'}{\{x^2\}'}&=& \lim_{x\to 0}\left(\frac{-\sin x}{\cos x}\cdot\frac{1}{2x}\right)\\ &=& \lim_{x\to 0}\left(\frac{\sin x}{x}\cdot\frac{-1}{2\cos x}\right)\\ &=& -\frac{1}{2}. \end{eqnarray} \]
Therefore,
\[\lim_{x\to 0}\frac{\log(\cos x)}{x^2} = -\frac{1}{2}.\]
□

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.
Remark. The same corollary holds if \(f(x)\) and \(g(x)\) are continuous functions on \((-\infty, b)\) and the limits \(\lim_{x\to \infty}\) are replaced with \(\lim_{x \to -\infty}\). □
Example. Let us find \(\lim_{x\to\infty}x^{\frac{1}{x}}\).
Let \(f(x) = x^{\frac{1}{x}}\) so that \(\log f(x) = \frac{\log x}{x}\). Now consider \(\lim_{x\to\infty}\log f(x)\).
  1. \(\lim_{x\to\infty}\log x = \infty\) and \(\lim_{x\to\infty}x = \infty\).
  2. \((x)' = 1 \neq 0\).
  3. \[\lim_{x\to\infty}\frac{(\log x)'}{(x)'} = \lim_{x\to\infty}\frac{1}{x} = 0.\]
Thus,
\[\lim_{x\to\infty}\frac{\log x}{x} = 0.\]
Noting that \(f(x) = e^{\log f(x)}\) and that the exponential function is continuous, we find
\[\lim_{x\to\infty}x^{\frac{1}{x}} = \lim_{x\to\infty}e^{\frac{\log x}{x}} = e^0 = 1.\]
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