Editing L'Hôpital's rule (section)

== Complications ==
Sometimes L'Hôpital's rule does not reduce to an obvious limit in a finite number of steps, unless some intermediate simplifications are applied. Examples include the following:

* Two applications can lead to a return to the original expression that was to be evaluated: <math display="block">
\lim_{x\to\infty} \frac{e^x+e^{-x}}{e^x-e^{-x}}
\ \stackrel{\mathrm{H}}{=}\  \lim_{x\to\infty} \frac{e^x-e^{-x}}{e^x+e^{-x}}
\ \stackrel{\mathrm{H}}{=}\  \lim_{x\to\infty} \frac{e^x+e^{-x}}{e^x-e^{-x}}
\ \stackrel{\mathrm{H}}{=}\  \cdots .
</math> This situation can be dealt with by substituting <math>y=e^x</math> and noting that {{math|''y''}} goes to infinity as {{math|''x''}} goes to infinity; with this substitution, this problem can be solved with a single application of the rule: <math display="block">
\lim_{x\to\infty} \frac{e^x+e^{-x}}{e^x-e^{-x}}
= \lim_{y\to\infty} \frac{y+y^{-1}}{y-y^{-1}}
\ \stackrel{\mathrm{H}}{=}\  \lim_{y\to\infty} \frac{1-y^{-2}}{1+y^{-2}}
= \frac{1}{1} = 1.
</math> Alternatively, the numerator and denominator can both be multiplied by <math>e^x,</math> at which point L'Hôpital's rule can immediately be applied successfully:<ref>Multiplying by <math>e^{-x}</math> instead yields a solution to the limit without need for l'Hôpital's rule.</ref> <math display="block">
\lim_{x\to\infty} \frac{e^x+e^{-x}}{e^x-e^{-x}}
= \lim_{x\to\infty} \frac{e^{2x} + 1}{e^{2x} - 1}
\ \stackrel{\mathrm{H}}{=}\  \lim_{x\to\infty} \frac{2e^{2x}}{2e^{2x}}
= 1.</math>
*An arbitrarily large number of applications may never lead to an answer even without repeating:<math display="block">
\lim_{x\to\infty} \frac{x^\frac1{2}+x^{-\frac1{2}}}{x^\frac1{2}-x^{-\frac1{2}}}
\ \stackrel{\mathrm{H}}{=}\  \lim_{x\to\infty} \frac{\frac1{2}x^{-\frac1{2}}-\frac{1}{2}x^{-\frac3{2}}}{\frac1{2}x^{-\frac1{2}}+\frac1{2}x^{-\frac3{2}}}
\ \stackrel{\mathrm{H}}{=}\  \lim_{x\to\infty} \frac{-\frac1{4}x^{-\frac3{2}}+\frac3{4}x^{-\frac5{2}}}{-\frac1{4}x^{-\frac3{2}}-\frac3{4}x^{-\frac5{2}}}
\ \stackrel{\mathrm{H}}{=}\  \cdots .</math>This situation too can be dealt with by a transformation of variables, in this case <math>y = \sqrt{x}</math>: <math display="block">
\lim_{x\to\infty} \frac{x^\frac1{2}+x^{-\frac1{2}}}{x^\frac1{2}-x^{-\frac1{2}}}
= \lim_{y\to\infty} \frac{y+y^{-1}}{y-y^{-1}}
\ \stackrel{\mathrm{H}}{=}\  \lim_{y\to\infty} \frac{1-y^{-2}}{1+y^{-2}}
= \frac1{1}
= 1.
</math> Again, an alternative approach is to multiply numerator and denominator by <math>x^{1/2}</math> before applying L'Hôpital's rule: <math display="block">
\lim_{x\to\infty} \frac{x^\frac{1}{2}+x^{-\frac{1}{2}}}{x^\frac{1}{2}-x^{-\frac{1}{2}}}
= \lim_{x\to\infty} \frac{x+1}{x-1}
\ \stackrel{\mathrm{H}}{=}\  \lim_{x\to\infty} \frac{1}{1}
= 1.</math>

A common logical fallacy is to use L'Hôpital's rule to prove the value of a derivative by computing the limit of a [[Derivative#Definition via difference quotients|difference quotient]]. Since applying l'Hôpital requires knowing the relevant derivatives, this amounts to [[circular reasoning]] or [[begging the question]], assuming what is to be proved. For example, consider the proof of the derivative formula for [[Derivative#Computing the derivative|powers of ''x'']]:

:<math>\lim_{h\to 0}\frac{(x+h)^n-x^n}{h}=nx^{n-1}.</math>

Applying L'Hôpital's rule and finding the derivatives with respect to {{math|''h''}} yields 
{{math|''nx''<sup>''n''−1</sup>}} as expected, but this computation requires the use of the very formula that is being proven. Similarly, to prove <math>\lim_{x\to 0}\frac{\sin(x)}{x}=1</math>, applying L'Hôpital requires knowing the derivative of <math>\sin(x)</math> at <math>x=0</math>, which amounts to calculating <math>\lim_{h\to 0}\frac{\sin(h)}{h}</math> in the first place; a valid proof requires a different method such as the [[squeeze theorem]].