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Subsection 3.7.3 Variations

Theorem 3.7.2 is the basic form of L'Hôpital's rule, but there are also many variations. Here are a bunch of them.

Subsubsection 3.7.3.1 Limits at \(\pm \infty\)

L'Hôpital's rule also applies when the limit of \(x \to a\) is replaced by \(\lim\limits_{x\rightarrow a+}\) or by \(\lim\limits_{x\rightarrow a-}\) or by \(\lim\limits_{x\rightarrow +\infty}\) or by \(\lim\limits_{x\rightarrow -\infty}\text{.}\)

We can justify adapting the rule to the limits to \(\pm \infty\) via the following reasoning

\begin{align*} \lim_{x\to \infty} \frac{f(x)}{g(x)} &= \lim_{y \to 0^+} \frac{ f(1/y) }{ g(1/y) } & \text{substitute } x=1/y\\ &= \lim_{y \to 0^+} \frac{ -\frac{1}{y^2} f'(1/y) } { -\frac{1}{y^2}g'(1/y)}, \end{align*}

where we have used l'Hôpital's rule (assuming this limit exists) and the fact that \(\diff{}{y} f(1/y) = -\frac{1}{y^2} f'(1/y)\) (and similarly for \(g\)). Cleaning this up and substituting \(y=1/x\) gives the required result:

\begin{align*} \lim_{x\to \infty} \frac{f(x)}{g(x)} & =\lim_{y \to 0^+} \frac{f'(1/y)} {g'(1/y)} =\lim_{x\to\infty} \frac{f'(x)}{g'(x)}. \end{align*}

Consider the limit

\begin{gather*} \lim_{x\to\infty} \frac{\arctan x - \frac{\pi}{2}}{ \frac{1}{x} } \end{gather*}

Both numerator and denominator go to \(0\) as \(x \to\infty\text{,}\) so this is an \(\frac00\) indeterminate form. We find

\begin{gather*} \lim_{x\rightarrow+\infty}\underbrace{\frac{\arctan x-\frac{\pi}{2}} {\frac{1}{x}}}_{\atp {\mathrm{num}\rightarrow 0} {\mathrm{den}\rightarrow 0}} =\lim_{x\rightarrow +\infty}\frac{\frac{1}{1+x^2}} {-\frac{1}{x^2}} =-\lim_{x\rightarrow +\infty}\underbrace{\frac{1}{1+\frac{1}{x^2}}}_{\atp {\mathrm{num}\rightarrow 1} {\mathrm{den}\rightarrow 1}} =-1 \end{gather*}

We have applied L'Hôpital's rule with

\begin{align*} f(x) &= \arctan x -\frac{\pi}{2}\qquad& g(x) &=\frac{1}{x}\\ f'(x)&=\frac{1}{1+x^2}& g'(x)&=-\frac{1}{x^2} \end{align*}

Subsubsection 3.7.3.2 \(\frac{\infty}{\infty}\) indeterminate form

L'Hôpital's rule also applies when \(\lim\limits_{x\rightarrow a}f(x)=0\text{,}\) \(\lim\limits_{x\rightarrow a}g(x)=0\) is replaced by \(\lim\limits_{x\rightarrow a}f(x)=\pm\infty\text{,}\) \(\lim\limits_{x\rightarrow a}g(x)=\pm\infty\text{.}\)

Consider the limit

\begin{gather*} \lim_{x \to \infty} \frac{\log x}{x} \end{gather*}

The numerator and denominator both blow up towards infinity so this is an \(\frac\infty\infty\) indeterminate form. An application of l'Hôpital's rule gives

\begin{align*} \lim_{x \to \infty} \underbrace{\frac{\log x}{x}}_{\atp {\mathrm{num}\rightarrow \infty}{\mathrm{den}\rightarrow \infty}} &= \lim_{x \to \infty} \frac{1/x}{1}\\ &= \lim_{x \to \infty} \frac{1}{x} = 0 \end{align*}

Consider the limit

\begin{gather*} \lim_{x \to \infty} \frac{5x^2+3x-3}{x^2+1} \end{gather*}

Then by two applications of l'Hôpital's rule we get

\begin{align*} \lim_{x \to \infty} \underbrace{\frac{5x^2+3x-3}{x^2+1}}_{\atp {\mathrm{num}\rightarrow \infty}{\mathrm{den}\rightarrow \infty}} &= \lim_{x \to \infty} \underbrace{\frac{10x+3}{2x}}_{\atp {\mathrm{num}\rightarrow \infty}{\mathrm{den}\rightarrow \infty}} = \lim_{x \to \infty} \frac{10}{2} = 5. \end{align*}

Compute the limit

\begin{gather*} \lim_{x\rightarrow 0+} \frac{\log x}{\tan\big(\frac{\pi}{2}-x\big)} \end{gather*}

We can compute this using l'Hôpital's rule twice:

\begin{align*} \lim_{x\rightarrow 0+}\underbrace{\frac{\log x} {\tan\big(\frac{\pi}{2}-x\big)}}_{\atp {\mathrm{num}\rightarrow -\infty} {\mathrm{den}\rightarrow +\infty}} & =\lim_{x\rightarrow 0+}\frac{\frac{1}{x}} {-\sec^2(\frac{\pi}{2}-x)} =-\lim_{x\rightarrow 0+}\underbrace{\frac{\cos^2(\frac{\pi}{2}-x)}{x}}_{\atp {\mathrm{num}\rightarrow 0} {\mathrm{den}\rightarrow 0}}\\ &=-\lim_{x\rightarrow 0+}\underbrace{\frac {2\cos(\frac{\pi}{2}-x)\sin(\frac{\pi}{2}-x)}{1}}_{\atp {\mathrm{num}\rightarrow 0} {\mathrm{den}\rightarrow 1}} =0 \end{align*}

The first application of L'Hôpital's was with

\begin{align*} f(x) &= \log x & g(x) &=\tan\Big(\frac{\pi}{2}-x\Big)\\ f'(x)&=\frac{1}{x}\qquad& g'(x)&=-\sec^2\Big(\frac{\pi}{2}-x\Big) \end{align*}

and the second time with

\begin{align*} f(x) &= \cos^2\Big(\frac{\pi}{2}-x\Big) & g(x) &=x\\ f'(x)&=2\cos\Big(\frac{\pi}{2}-x\Big) \Big[-\sin\Big(\frac{\pi}{2}-x\Big)\Big](-1)\qquad& g'(x)&=1 \end{align*}

Sometimes things don't quite work out as we would like and l'Hôpital's rule can get stuck in a loop. Remember to think about the problem before you apply any rule.

Consider the limit

\begin{gather*} \lim_{x\to\infty} \frac{ e^x + e^{-x} }{e^x - e^{-x}} \end{gather*}

Clearly both numerator and denominator go to \(\infty\text{,}\) so we have a \(\frac\infty\infty\) indeterminate form. Naively applying l'Hôpital's rule gives

\begin{align*} \lim_{x\to\infty} \frac{ e^x + e^{-x} }{e^x - e^{-x}} &= \lim_{x\to\infty} \frac{ e^x - e^{-x} }{e^x + e^{-x}} \end{align*}

which is again a \(\frac\infty\infty\) indeterminate form. So apply l'Hôpital's rule again:

\begin{align*} \lim_{x\to\infty} \frac{ e^x - e^{-x} }{e^x + e^{-x}} &= \lim_{x\to\infty} \frac{ e^x + e^{-x} }{e^x - e^{-x}} \end{align*}

which is right back where we started!

The correct approach to such a limit is to apply the methods we learned in Chapter 1 and rewrite

\begin{align*} \frac{e^x+e^{-x}}{e^x-e^{-x}} &= \frac{e^x(1+e^{-2x})}{e^x(1-e^{-2x})} = \frac{1+e^{-2x}}{1-e^{-2x}} \end{align*}

and then take the limit.

A similar sort of l'Hôpital-rule-loop will occur if you naively apply l'Hôpital's rule to the limit

\begin{gather*} \lim_{x\to\infty} \frac{\sqrt{4x^2+1}}{5x-1} \end{gather*}

which appeared in Example 1.5.6.

Subsubsection 3.7.3.3 Optional — Proof of l'Hôpital's Rule for \(\frac\infty\infty\)

We can justify this generalisation of l'Hôpital's rule with some careful manipulations. Since the derivatives \(f',g'\) exist in some interval around \(a\text{,}\) we know that \(f,g\) are continuous in some interval around \(a\text{;}\) let \(x,t\) be points inside that interval. Now rewrite  4 This is quite a clever argument, but it is not immediately obvious why one rewrites things this way. After the fact it becomes clear that it is done to massage the expression into the form where we can apply the generalised mean-value theorem (Theorem 3.4.38).

\begin{align*} \frac{f(x)}{g(x)} &= \frac{f(x)}{g(x)} +\underbrace{\left( \frac{f(t)}{g(x)} - \frac{f(t)}{g(x)}\right)}_{=0} + \underbrace{\left(\frac{f(x)-f(t)}{g(x)-g(t)} - \frac{f(x)-f(t)}{g(x)-g(t)}\right)}_{=0}\\ &= \underbrace{\frac{f(x)-f(t)}{g(x)-g(t)}}_\text{ready for GMVT} + \frac{f(t)}{g(x)} + \underbrace{\left( \frac{f(x)}{g(x)} - \frac{f(t)}{g(x)} - \frac{f(x)-f(t)}{g(x)-g(t)} \right)}_\text{we can clean it up}\\ &= \frac{f(x)-f(t)}{g(x)-g(t)} + \frac{f(t)}{g(x)} + \left( \frac{f(x)-f(t)}{g(x)} - \frac{f(x)-f(t)}{g(x)-g(t)} \right)\\ &= \frac{f(x)-f(t)}{g(x)-g(t)} + \frac{f(t)}{g(x)} + \left( \frac{1}{g(x)} - \frac{1}{g(x)-g(t)} \right)\cdot (f(x)-f(t))\\ &= \frac{f(x)-f(t)}{g(x)-g(t)} + \frac{f(t)}{g(x)} + \left( \frac{g(x)-g(t) - g(x)}{g(x)(g(x)-g(t))} \right)\cdot (f(x)-f(t))\\ &= \underbrace{\frac{f(x)-f(t)}{g(x)-g(t)}}_\text{ready for GMVT} + \frac{f(t)}{g(x)} - \frac{g(t)}{g(x)}\cdot \underbrace{\frac{f(x)-f(t)}{g(x)-g(t)}}_\text{ready for GMVT} \end{align*}

Oof! Now the generalised mean-value theorem (Theorem 3.4.38) tells us there is a \(c\) between \(x\) and \(t\) so that

\begin{align*} \frac{f(x)-f(t)}{g(x)-g(t)} &= \frac{f'(c)}{g'(c)} \end{align*}

Now substitute this into the large expression we derived above:

\begin{align*} \frac{f(x)}{g(x)} &= \frac{f'(c)}{g'(c)} + \frac{1}{g(x)}\left(f(t) - \frac{f'(c)}{g'(c)}\cdot g(t) \right) \end{align*}

At first glance this does not appear so useful, however if we fix \(t\) and take the limit as \(x \rightarrow a\text{,}\) then it becomes

\begin{align*} \lim_{x\to a}\frac{f(x)}{g(x)} &= \lim_{x\to a} \frac{f'(c)}{g'(c)} + \lim_{x\to a} \frac{1}{g(x)}\left(f(t) - \frac{f'(c)}{g'(c)}\cdot g(t) \right)\\ \end{align*}

Since \(g(x) \to \infty\) as \(x \to a\text{,}\) this last term goes to zero

\begin{align*} &= \lim_{x\to a} \frac{f'(c)}{g'(c)} + 0 \end{align*}

Now take the limit as \(t\to a\text{.}\) The left-hand side is unchanged since it is independent of \(t\text{.}\) The right-hand side, however, does change; the number \(c\) is trapped between \(x\) and \(t\text{.}\) Since we have already taken the limit \(x\to a\text{,}\) so when we take the limit \(t \to a\text{,}\) we are effectively taking the limit \(c \to a\text{.}\) Hence

\begin{align*} \lim_{x\to a}\frac{f(x)}{g(x)} &= \lim_{c\to a} \frac{f'(c)}{g'(c)} \end{align*}

which is the desired result.

Subsubsection 3.7.3.4 \(0\cdot\infty\) indeterminate form

When \(\ds \lim_{x\to a}f(x) = 0\) and \(\ds \lim_{x\to a} g(x) = \infty\text{.}\) We can use a little algebra to manipulate this into either a \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) form:

\begin{align*} \lim_{x \to a} \frac{f(x)}{1/g(x)} && \lim_{x \to a} \frac{g(x)}{1/f(x)} \end{align*}

Consider the limit

\begin{gather*} \lim_{x\to 0^+} x \cdot \log x \end{gather*}

Here the function \(f(x)=x\) goes to zero, while \(g(x)=\log x\) goes to \(-\infty\text{.}\) If we rewrite this as the fraction

\begin{align*} x \cdot \log x &= \frac{\log x}{1/x} \end{align*}

then the \(0 \cdot \infty\) form has become an \(\frac\infty\infty\) form.

The result is then

\begin{align*} \lim_{x\rightarrow 0+}\underbrace{x}_{\rightarrow 0} \underbrace{\log x}_{\rightarrow-\infty} & =\lim_{x\rightarrow 0+}\underbrace{\frac{\log x} {\frac{1}{x}}}_{\atp {\mathrm{num}\rightarrow-\infty} {\mathrm{den}\rightarrow\infty}} =\lim_{x\rightarrow 0+}\frac{\frac{1}{x}}{-\frac{1}{x^2}} =-\lim_{x\rightarrow 0+}x =0 \end{align*}

In this example we'll evaluate \(\lim\limits_{x\rightarrow +\infty} x^n e^{-x}\text{,}\) for all natural numbers \(n\text{.}\) We'll start with \(n=1\) and \(n=2\) and then, using what we have learned from those cases, move on to general \(n\text{.}\)

\begin{align*} \lim_{x\rightarrow +\infty}\underbrace{x}_{\rightarrow \infty} \underbrace{e^{-x}}_{\rightarrow 0} & =\lim_{x\rightarrow +\infty}\underbrace{\frac{x} {e^x}}_{\atp {\mathrm{num}\rightarrow+\infty} {\mathrm{den}\rightarrow+\infty}} =\lim_{x\rightarrow +\infty}\underbrace{\frac{1} {e^x}}_{\atp {\mathrm{num}\rightarrow 1} {\mathrm{den}\rightarrow+\infty}} =\lim_{x\rightarrow +\infty}e^{-x} =0 \end{align*}

Applying l'Hôpital twice,

\begin{align*} \lim_{x\rightarrow +\infty}\underbrace{x^2}_{\rightarrow \infty} \underbrace{e^{-x}}_{\rightarrow 0} & =\lim_{x\rightarrow +\infty}\underbrace{\frac{x^2} {e^x}}_{\atp {\mathrm{num}\rightarrow+\infty} {\mathrm{den}\rightarrow+\infty}} =\lim_{x\rightarrow +\infty}\underbrace{\frac{2x} {e^x}}_{\atp {\mathrm{num}\rightarrow \infty} {\mathrm{den}\rightarrow+\infty}} =\lim_{x\rightarrow +\infty}\underbrace{\frac{2} {e^x}}_{\atp {\mathrm{num}\rightarrow 2} {\mathrm{den}\rightarrow+\infty}} =\lim_{x\rightarrow +\infty}2e^{-x}\\ \amp=0 \end{align*}

Indeed, for any natural number \(n\text{,}\) applying l'Hôpital \(n\) times gives

\begin{align*} \lim_{x\rightarrow +\infty}\underbrace{x^n}_{\rightarrow \infty} \underbrace{e^{-x}}_{\rightarrow 0} & =\lim_{x\rightarrow +\infty}\hskip-5pt\underbrace{\frac{x^n} {e^x}}_{\atp {\mathrm{num}\rightarrow+\infty} {\mathrm{den}\rightarrow+\infty}}\\ & =\lim_{x\rightarrow +\infty}\underbrace{\frac{nx^{n-1}} {e^x}}_{\atp {\mathrm{num}\rightarrow \infty} {\mathrm{den}\rightarrow+\infty}}\\ & =\lim_{x\rightarrow +\infty}\underbrace{\frac{n(n-1)x^{n-2}} {e^x}}_{\atp {\mathrm{num}\rightarrow \infty} {\mathrm{den}\rightarrow+\infty}}\\ &=\cdots =\lim_{x\rightarrow +\infty}\underbrace{\frac{n!} {e^x}}_{\atp {\mathrm{num}\rightarrow n!} {\mathrm{den}\rightarrow+\infty}} =0 \end{align*}

Subsubsection 3.7.3.5 \(\infty-\infty\) indeterminate form

When \(\ds \lim_{x\to a}f(x) = \infty\) and \(\ds \lim_{x\to a} g(x) = \infty\text{.}\) We rewrite the difference as a fraction using a common denominator

\begin{align*} f(x) - g(x) &= \frac{h(x)}{\ell(x)} \end{align*}

which is then a \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) form.

Consider the limit

\begin{gather*} \lim_{x\to \frac{\pi}{2}^-} \left( \sec x - \tan x\right) \end{gather*}

Since the limit of both \(\sec x\) and \(\tan x\) is \(+\infty\) as \(x \to \frac{\pi}{2}^-\text{,}\) this is an \(\infty-\infty\) indeterminate form. However we can rewrite this as

\begin{align*} \sec x - \tan x &= \frac{1}{\cos x} - \frac{\sin x}{\cos x} = \frac{1-\sin x}{\cos x} \end{align*}

which is then a \(\frac00\) indeterminate form. This then gives

\begin{align*} \lim_{x\rightarrow \frac{\pi}{2}^-}\Big( \underbrace{\sec x}_{\rightarrow +\infty} - \underbrace{\tan x}_{\rightarrow+\infty}\Big) & =\lim_{x\rightarrow \frac{\pi}{2}^-}\underbrace{\frac{1-\sin x} {\cos x}}_{\atp {\mathrm{num}\rightarrow 0} {\mathrm{den}\rightarrow 0}} =\lim_{x\rightarrow \frac{\pi}{2}^-}\underbrace{\frac{-\cos x} {-\sin x}}_{\atp {\mathrm{num}\rightarrow 0} {\mathrm{den}\rightarrow -1}} =0 \end{align*}

In the last example, Example 3.7.17, we converted an \(\infty-\infty\) indeterminate form into a \(\frac{0}{0}\) indeterminate form by exploiting the fact that the two terms, \(\sec x\) and \(\tan x\text{,}\) in the \(\infty-\infty\) indeterminate form shared a common denominator, namely \(\cos x\text{.}\) In the “real world” that will, of course, almost never happen. However as the next couple of examples show, you can often massage these expressions into suitable forms.

Here is another, much more complicated, example, where it doesn't happen.

In this example, we evaluate the \(\infty-\infty\) indeterminate form

\begin{gather*} \lim_{x\rightarrow 0}\Big( \underbrace{\frac{1}{x}}_{\rightarrow \pm\infty} - \underbrace{\frac{1}{\log(1+x)}}_{\rightarrow\pm\infty}\Big) \end{gather*}

We convert it into a \(\frac{0}{0}\) indeterminate form simply by putting the two fractions, \(\frac{1}{x}\) and \(\frac{1}{\log(1+x)}\) over a common denominator.

\begin{equation*} \lim_{x\rightarrow 0}\Big( \underbrace{\frac{1}{x}}_{\rightarrow \pm\infty} - \underbrace{\frac{1}{\log(1+x)}}_{\rightarrow\pm\infty}\Big) =\lim_{x\rightarrow 0}\underbrace{\frac{\log(1+x)-x}{x\log(1+x)}}_{\atp {\mathrm{num}\rightarrow 0} {\mathrm{den}\rightarrow 0}} \tag{E1} \end{equation*}

Now we apply L'Hôpital's rule, and simplify

\begin{align*} \lim_{x\rightarrow 0}\underbrace{\frac{\log(1+x)-x}{x\log(1+x)}}_{\atp {\mathrm{num}\rightarrow 0} {\mathrm{den}\rightarrow 0}} &=\lim_{x\rightarrow 0}\frac{\frac{1}{1+x}-1} {\log(1+x)+\frac{x}{1+x}}\\ \amp=\lim_{x\rightarrow 0}\frac{1-(1+x)} {(1+x)\log(1+x)+x}\\ &=-\lim_{x\rightarrow 0}\underbrace{\frac{x} {(1+x)\log(1+x)+x}}_{\atp {\mathrm{num}\rightarrow 0} {\mathrm{den}\rightarrow 1\times 0+0=0}} \tag{E2} \end{align*}

Then we apply L'Hôpital's rule a second time

\begin{align*} -\lim_{x\rightarrow 0}\underbrace{\frac{x} {(1+x)\log(1+x)+x}}_{\atp {\mathrm{num}\rightarrow 0} {\mathrm{den}\rightarrow 1\times 0+0=0}} \amp=-\lim_{x\rightarrow 0}\underbrace{\frac{1} {\log(1+x)+\frac{1+x}{1+x}+1}}_{\atp {\mathrm{num}\rightarrow 1} {\mathrm{den}\rightarrow 0+1+1=2}}\\ \amp=-\frac{1}{2} \tag{E3} \end{align*}

Combining (E1), (E2) and (E3) gives our final answer

\begin{gather*} \lim_{x\rightarrow 0}\Big(\frac{1}{x} - \frac{1}{\log(1+x)}\Big) =-\frac{1}{2} \end{gather*}

The following example can be done by l'Hôpital's rule, but it is actually far simpler to multiply by the conjugate and take the limit using the tools of Chapter 1.

Consider the limit

\begin{gather*} \lim_{x\to \infty} \sqrt{x^2+4x}-\sqrt{x^2-3x} \end{gather*}

Neither term is a fraction, but we can write

\begin{align*} \sqrt{x^2+4x}-\sqrt{x^2-3x} &= x\sqrt{1+4/x}-x\sqrt{1-3/x} & \text{assuming } x \gt 0\\ &= x \left( \sqrt{1+4/x}-\sqrt{1-3/x} \right)\\ &= \frac{\sqrt{1+4/x}-\sqrt{1-3/x}}{1/x} \end{align*}

which is now a \(\frac00\) form with \(f(x)=\sqrt{1+4/x}-\sqrt{1-3/x}\) and \(g(x)=1/x\text{.}\) Then

\begin{align*} f'(x) &= \frac{-4/x^2}{2\sqrt{1+4/x}} - \frac{3/x^2}{2\sqrt{1-3/x}} & g'(x) &= - \frac{1}{x^2} \end{align*}

Hence

\begin{align*} \frac{f'(x)}{g'(x)} &= \frac{4}{2\sqrt{1+4/x}} + \frac{3}{\sqrt{1-3/x}} \end{align*}

And so in the limit as \(x\to \infty\)

\begin{align*} \lim_{x\to \infty} \frac{f'(x)}{g'(x)} &= \frac{4}{2}+\frac{3}{2} = \frac{7}{2} \end{align*}

and so our original limit is also \(7/2\text{.}\)

By comparison, if we multiply by the conjugate we have

\begin{align*} \sqrt{x^2\!+\!4x}-\sqrt{x^2\!-\!3x} &= \left( \sqrt{x^2\!+\!4x}-\sqrt{x^2\!-\!3x}\right) \cdot \frac{\sqrt{x^2\!+\!4x}+\sqrt{x^2\!-\!3x}}{\sqrt{x^2\!+\!4x}+\sqrt{x^2\!-\!3x}}\\ &= \frac{ x^2+4x - (x^2-3x)}{\sqrt{x^2+4x}+\sqrt{x^2-3x}}\\ &= \frac{ 7x}{\sqrt{x^2+4x}+\sqrt{x^2-3x}}\\ &= \frac{7}{\sqrt{1+4/x}+\sqrt{1-3/x}} \qquad \text{assuming } x \gt 0 \end{align*}

Now taking the limit as \(x\to\infty\) gives \(7/2\) as required. Just because we know l'Hôpital's rule, it does not mean we should use it everywhere it might be applied.

Subsubsection 3.7.3.6 \(1^\infty\) indeterminate form

We can use l'Hôpital's rule on limits of the form

\begin{align*} \lim_{x\to a} f(x)^{g(x)} & \text{ with }\\ \lim_{x\to a} f(x) &= 1 & \text{ and } && \lim_{x \to a} g(x) &= \infty \end{align*}

by considering the logarithm of the limit  5 We are using the fact that the logarithm is a continuous function and Theorem 1.6.10. :

\begin{align*} \log\left( \lim_{x\to a} f(x)^{g(x)} \right) &= \lim_{x\to a} \log\left( f(x)^{g(x)} \right) = \lim_{x\to a} \log\left( f(x) \right) \cdot g(x) \end{align*}

which is now an \(0 \cdot \infty\) form. This can be further transformed into a \(\frac00\) or \(\frac\infty\infty\) form:

\begin{align*} \log\left( \lim_{x\to a} f(x)^{g(x)} \right) &=\lim_{x\to a} \log\left( f(x) \right) \cdot g(x)\\ &= \lim_{x\to a} \frac{\log\left( f(x) \right)}{1/g(x)}. \end{align*}

The following limit appears quite naturally when considering systems which display exponential growth or decay.

\begin{gather*} \lim_{x\rightarrow 0}(1+x)^{\frac{a}{x}} \qquad\text{with the constant } a\ne 0 \end{gather*}

Since \((1+x) \to 1\) and \(a/x \to \infty\) this is an \(1^\infty\) indeterminate form.

By considering its logarithm we have

\begin{align*} \log\left( \lim_{x\to 0}(1+x)^{\frac{a}{x}}\right) &= \lim_{x\to 0}\log\left( (1+x)^{\frac{a}{x}}\right)\\ &= \lim_{x\to 0} \frac{a}{x} \log(1+x)\\ &= \lim_{x\to 0} \frac{a\log (1+x)}{x} \end{align*}

which is now a \(\frac00\) form. Applying l'Hôpital's rule gives

\begin{gather*} \lim_{x\rightarrow 0}\underbrace{\frac{a\log(1+x)}{x}}_{\atp {\mathrm{num}\rightarrow 0} {\mathrm{den}\rightarrow 0}} =\lim_{x\rightarrow 0}\underbrace{\frac{\frac{a}{1+x}}{1}}_{\atp {\mathrm{num}\rightarrow a} {\mathrm{den}\rightarrow 1}} =a \end{gather*}

Since \((1 + x)^{a/x} = \exp\left[\log\Big((1 + x)^{a/x}\Big)\right]\) and the exponential function is continuous, our original limit is \(e^a\text{.}\)

Here is a more complicated example of a \(1^\infty\) indeterminate form.

In the limit

\begin{gather*} \lim_{x\rightarrow 0}\Big(\frac{\sin x}{x}\Big)^{\frac{1}{x^2}} \end{gather*}

the base, \(\frac{\sin x}{x}\text{,}\) converges to \(1\) (see Example 3.7.3) and the exponent, \(\frac{1}{x^2}\text{,}\) goes to \(\infty\text{.}\) But if we take logarithms then

\begin{gather*} \log \Big(\frac{\sin x}{x}\Big)^{\frac{1}{x^2}} = \frac{\log\frac{\sin x}{x}}{x^2} \end{gather*}

then, in the limit \(x\rightarrow 0\text{,}\) we have a \(\frac{0}{0}\) indeterminate form. One application of l'Hôpital's rule gives

\begin{align*} \lim_{x\rightarrow 0}\underbrace{\frac{\log\frac{\sin x}{x}}{x^2}}_{\atp {\mathrm{num}\rightarrow 0} {\mathrm{den}\rightarrow 0}} & =\lim_{x\rightarrow 0} \frac {\frac{x}{\sin x}\frac{x\cos x -\sin x}{x^2} }{2x} =\lim_{x\rightarrow 0} \frac{ \frac{x\cos x -\sin x}{x\sin x} }{2x} =\lim_{x\rightarrow 0} \frac{x\cos x -\sin x}{2x^2\sin x} \end{align*}

which is another \(\frac00\) form. Applying l'Hôpital's rule again gives:

\begin{align*} \lim_{x\rightarrow 0}\underbrace{ \frac{x\cos x -\sin x}{2x^2\sin x} }_{\atp {\mathrm{num}\rightarrow 0} {\mathrm{den}\rightarrow 0}} &=\lim_{x\rightarrow 0} \frac{\cos x -x\sin x-\cos x}{4x\sin x+2x^2\cos x}\\ &=-\lim_{x\rightarrow 0} \frac{x\sin x}{4x\sin x+2x^2\cos x} =-\lim_{x\rightarrow 0} \frac{\sin x}{4\sin x+2x\cos x} \end{align*}

which is yet another \(\frac00\) form. Once more with l'Hôpital's rule:

\begin{align*} -\lim_{x\rightarrow 0}\underbrace{ \frac{\sin x}{4\sin x+2x\cos x} }_{\atp {\mathrm{num}\rightarrow 0} {\mathrm{den}\rightarrow 0}} &=-\lim_{x\rightarrow 0}\underbrace{ \frac{\cos x}{4\cos x+2\cos x-2x\sin x} }_{\atp {\mathrm{num}\rightarrow 1} {\mathrm{den}\rightarrow 6}}\\ &=-\frac{1}{6} \end{align*}

Oof! We have just shown that the logarithm of our original limit is \(-\frac{1}{6}\text{.}\) Hence the original limit itself is \(e^{-1/6}\text{.}\)

This was quite a complicated example. However it does illustrate the importance of cleaning up your algebraic expressions. This will both reduce the amount of work you have to do and will also reduce the number of errors you make.

Subsubsection 3.7.3.7 \(0^0\)indeterminate form

Like the \(1^\infty\) form, this can be treated by considering its logarithm.

For example, in the limit

\begin{gather*} \lim_{x\rightarrow 0+}x^x \end{gather*}

both the base, \(x\text{,}\) and the exponent, also \(x\text{,}\) go to zero. But if we consider the logarithm then we have

\begin{gather*} \log x^x = x\log x \end{gather*}

which is a \(0\cdot\infty\) indeterminate form, which we already know how to treat. In fact, we already found, in Example 3.7.15, that

\begin{gather*} \lim_{x\rightarrow 0+} x\log x=0 \end{gather*}

Since the exponential is a continuous function

\begin{gather*} \lim_{x\rightarrow 0+}x^x =\lim_{x\rightarrow 0+}\exp\big(x\log x\big) =\exp\Big(\lim_{x\rightarrow 0+}x\log x\Big) =e^0 =1 \end{gather*}

Subsubsection 3.7.3.8 \(\infty^0\) indeterminate form

Again, we can treat this form by considering its logarithm.

For example, in the limit

\begin{gather*} \lim_{x\rightarrow +\infty}x^{\frac{1}{x}} \end{gather*}

the base, \(x\text{,}\) goes to infinity and the exponent, \(\frac{1}{x}\text{,}\) goes to zero. But if we take logarithms

\begin{gather*} \log x^{\frac{1}{x}} =\frac{\log x}{x} \end{gather*}

which is an \(\frac\infty\infty\) form, which we know how to treat.

\begin{gather*} \lim_{x\rightarrow +\infty}\underbrace{\frac{\log x}{x}}_{\atp {\mathrm{num}\rightarrow \infty} {\mathrm{den}\rightarrow \infty}} =\lim_{x\rightarrow +\infty}\underbrace{\frac{\frac{1}{x}}{1}}_{\atp {\mathrm{num}\rightarrow 0} {\mathrm{den}\rightarrow 1}} =0 \end{gather*}

Since the exponential is a continuous function

\begin{gather*} \lim_{x\rightarrow +\infty}x^{\frac{1}{x}} =\lim_{x\rightarrow +\infty}\exp\Big(\frac{\log x}{x}\Big) =\exp\Big(\lim_{x\rightarrow \infty}\frac{\log x}{x}\Big) =e^0 =1 \end{gather*}