Editing L'Hôpital's rule (section)

== Necessity of conditions: Counterexamples ==

All four conditions for l'Hôpital's rule are necessary:

# Indeterminacy of form: <math> \lim_{x \to c} f(x) = \lim_{x \to c} g(x) = 0 </math> or <math> \pm \infty </math> ; 
# Differentiability of functions: <math> f(x) </math> and <math> g(x) </math> are [[Differentiable function|differentiable]] on an open [[Interval (mathematics)|interval]] <math> \mathcal{I} </math> except possibly at the limit point <math> c </math> in <math> \mathcal{I} </math>; 
# Non-zero derivative of denominator: <math> g'(x) \ne 0 </math> for all <math> x </math> in <math> \mathcal{I} </math> with <math> x \ne c </math> ; 
# Existence of limit of the quotient of the derivatives: <math> \lim_{x \to c} \frac{f'(x)}{g'(x)} </math> exists.

Where one of the above conditions is not satisfied, l'Hôpital's rule is not valid in general, and its conclusion may be false in certain cases.

=== 1. Form is not indeterminate ===
The necessity of the first condition can be seen by considering the counterexample where the functions are <math> f(x) = x +1 </math> and <math> g(x) = 2x +1 </math> and the limit is <math> x \to 1 </math>.

The first condition is not satisfied for this counterexample because <math> \lim_{x \to 1} f(x) = \lim_{x \to 1} (x + 1) = (1) + 1 = 2 \neq 0 </math> and <math> \lim_{x \to 1} g(x) = \lim_{x \to 1} (2x + 1) = 2(1) + 1 = 3 \neq 0 </math>. This means that the form is not indeterminate.

The second and third conditions are satisfied by <math> f(x) </math> and <math> g(x) </math>. The fourth condition is also satisfied with

<math display="block"> \lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{(x+1)'}{(2x+1)'} 
= \lim_{x \to 1} \frac{1}{2} = \frac{1}{2}. </math>

But the conclusion fails, since <math display="block"> \lim_{x \to 1} \frac{f(x)}{g(x)} = \lim_{x \to 1} \frac{x+1}{2x+1} = \frac{ \lim_{x \to 1} (x+1) }{ \lim_{x \to 1} (2x+1) } = \frac{2}{3} \neq \frac{1}{2}. </math>
=== 2. Differentiability of functions ===
Differentiability of functions is a requirement because if a function is not differentiable, then the derivative of the function is not guaranteed to exist at each point in <math> \mathcal{I} </math>. The fact that <math> \mathcal{I} </math> is an open interval is grandfathered in from the hypothesis of the [[Cauchy's mean value theorem]]. The notable exception of the possibility of the functions being not differentiable at <math> c </math> exists because l'Hôpital's rule only requires the derivative to exist as the function approaches <math> c </math>; the derivative does not need to be taken at <math> c </math>. 

For example, let <math> f(x) = \begin{cases} \sin x, & x\neq0 \\ 1, & x=0 \end{cases} </math> , <math> g(x)=x </math>, and <math> c = 0 </math>. In this case, <math> f(x) </math> is not differentiable at <math> c </math>. However, since <math> f(x) </math> is differentiable everywhere except <math> c </math>, then <math> \lim_{x \to c}f'(x) </math> still exists. Thus, since 

<math> \lim_{x\to c} \frac{f(x)}{g(x)} = \frac{0}{0} </math> and <math> \lim_{x\to c} \frac{f'(x)}{g'(x)}  </math> exists, l'Hôpital's rule still holds.

=== 3. Derivative of denominator is zero ===
The necessity of the condition that <math>g'(x)\ne 0</math> near <math>c</math> can be seen by the following counterexample due to [[Otto Stolz]].<ref name="stolz">{{Cite journal |last=Stolz |first=Otto |author-link=Otto Stolz |year=1879 |title=Ueber die Grenzwerthe der Quotienten |trans-title=About the limits of quotients |url=https://resolver.sub.uni-goettingen.de/purl?235181684_0015{{!}}log39 |journal=[[Mathematische Annalen]] |language=German |volume=15 |issue=3–4 |pages=556–559 |doi=10.1007/bf02086277|s2cid=122473933 }}</ref> Let <math>f(x)=x+\sin x \cos x</math> and <math>g(x)=f(x)e^{\sin x}.</math> Then there is no limit for <math>f(x)/g(x)</math> as <math>x\to\infty.</math> However,

:<math>\begin{align}
\frac{f'(x)}{g'(x)} &= \frac{2\cos^2 x}{(2 \cos^2 x) e^{\sin x} + (x+\sin x \cos x) e^{\sin x} \cos x} \\
&= \frac{2\cos x}{2 \cos x +x+\sin x \cos x} e^{-\sin x},
\end{align}</math>

which tends to 0 as <math>x\to\infty</math>, although it is undefined at infinitely many points. Further examples of this type were found by [[Ralph P. Boas Jr.]]<ref name="boas">{{Cite journal |last=Boas Jr. |first=Ralph P.  |author-link=Ralph P. Boas Jr. |title=Counterexamples to L'Hopital's Rule |date=1986|journal=[[American Mathematical Monthly]] |volume=93 |issue=8 |pages=644–645 |jstor=2322330 |doi=10.1080/00029890.1986.11971912 }}</ref>

=== 4. Limit of derivatives does not exist ===
The requirement that the limit <math>\lim_{x\to c}\frac{f'(x)}{g'(x)}</math> exists is essential; if it does not exist, the original limit <math>\lim_{x\to c}\frac{f(x)}{g(x)}</math> may nevertheless exist. Indeed, as <math>x</math> approaches <math>c</math>, the functions <math>f</math> or <math>g</math> may exhibit many oscillations of small amplitude but steep slope, which do not affect <math>\lim_{x\to c}\frac{f(x)}{g(x)}</math> but do prevent the convergence of <math>\lim_{x\to c}\frac{f'(x)}{g'(x)}</math>. 

For example, if <math>f(x)=x+\sin(x)</math>, <math>g(x)=x</math> and <math>c=\infty</math>, then <math display="block">\frac{f'(x)}{g'(x)}=\frac{1+\cos(x)}{1},</math>which does not approach a limit since cosine oscillates infinitely between {{math|1}} and {{math|−1}}. But the ratio of the original functions does approach a limit, since the amplitude of the oscillations of <math>f</math> becomes small relative to <math>g</math>:

:<math>\lim_{x\to\infty}\frac{f(x)}{g(x)} 
= \lim_{x\to\infty}\left(\frac{x+\sin(x)}{x}\right) 
= \lim_{x\to\infty}\left(1+\frac{\sin(x)}{x}\right) 
= 1+0 = 1. </math>

In a case such as this, all that can be concluded is that

: <math> \liminf_{x \to c} \frac{f'(x)}{g'(x)} \leq \liminf_{x \to c} \frac{f(x)}{g(x)} \leq \limsup_{x \to c} \frac{f(x)}{g(x)} \leq \limsup_{x \to c} \frac{f'(x)}{g'(x)} ,</math>

so that if the limit of <math display="inline">\frac{f}{g}
</math> exists, then it must lie between the inferior and superior limits of <math display="inline">\frac{f'}{g'}
</math> .  In the example, 1 does indeed lie between 0 and 2.)

Note also that by the [[Contraposition|contrapositive]] form of the Rule, if <math>\lim_{x\to c}\frac{f(x)}{g(x)}</math> does not exist, then <math>\lim_{x\to c}\frac{f'(x)}{g'(x)}</math> also does not exist.