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I need to evaluate the following limit using l'Hospital's rule: $$\lim_{x\to 0}\dfrac{1-(\cos x)^{\sin x}}{x^3}$$

By doing one step, i get $$\lim_{x\to 0}\dfrac{-(\cos x)^{\sin x}[(\cos x) \ln(\cos x)-\frac{(\sin^2 x)}{\cos x}]}{3x^2}$$

If I did this correctly, I still need to use l'Hospital's rule again, but this seems too complicated for an exam question. Is there another, simpler way of doing this, but by still using L'Hospital's.

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  • $\begingroup$ i think this is not correct $\endgroup$
    – Dr. Sonnhard Graubner
    Jan 21, 2017 at 15:33
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    $\begingroup$ 1. Do they force you to use L'Hopital rule to solve this? 2. If they do, what is your question? Whether your first application of the rule is correct? (If they do not, let me mention that some other approaches require significantly less computations and allow to know what one is doing. To wit, I can tell you without even grabbing a pen and a piece of paper that the limit is $\color{red}{\frac12}\cdot\color{green}{1}$, because $1-\cos x\sim\color{red}{\frac12}x^2$ and $\sin x\sim\color{green}{1}\cdot x$ around $x=0$. So, in a way, L'H seems to be the worst approach there is to solve this.) $\endgroup$
    – Did
    Jan 21, 2017 at 15:37
  • $\begingroup$ the derivative of your numerator is given by $$-\cos ^{\sin (x)}(x) (\cos (x) \log (\cos (x))-\sin (x) \tan (x))$$ $\endgroup$
    – Dr. Sonnhard Graubner
    Jan 21, 2017 at 15:38
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    $\begingroup$ and the denominator $$3x^2$$ $\endgroup$
    – Dr. Sonnhard Graubner
    Jan 21, 2017 at 15:38
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    $\begingroup$ This is a prime example of how L'Hopital's Rule is not the be all do all of limit evaluation. $\endgroup$
    – Malcolm
    Jan 21, 2017 at 15:44

3 Answers 3

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There is a way of solving this using l'Hospital's rule.

$$\lim_{x\to 0}\dfrac{1-(\cos x)^{\sin x}}{x^3}=\lim_{x\to 0}\dfrac{1-(\cos x)^{\sin x}}{\sin^{3}{x}}$$

Apply the rule, we get:

$$\lim_{x\to 0}\dfrac{-(\cos x)^{\sin {x}}[(\cos x) \ln(\cos x)-\frac{(\sin^2 x)}{\cos x}]}{3\sin^2{x}\cos{x}}$$ $$\lim_{x\to 0}-\dfrac{-(\cos x)^{(\sin {x}-1)}[(\cos x) \ln(\cos x)-\frac{(\sin^2 x)}{\cos x}]}{3\sin^2{x}}$$ $$=\lim_{x\to 0}\dfrac{[(\cos x) \ln(\cos x)-\frac{(\sin^2 x)}{\cos x}]}{-3\sin^2{x}}$$ $$=\lim_{x\to 0}\dfrac{[(\cos^2 {x}) \ln(\cos x)-{(\sin^2 x)}]}{-3\sin^2{x}\cos{x}}$$ $$=\lim_{x\to 0}\dfrac{[(\cos^2 {x}) \ln(\cos x)-{(\sin^2 x)}]}{-3\sin^2{x}}$$ $$=\lim_{x\to 0}\dfrac{(\cos^2 {x}) \ln(\cos x)}{-3\sin^2{x}}+\lim_{x\to 0}\dfrac{{(\sin^2 x)}}{3\sin^2{x}}$$ $$=\lim_{x\to 0}\dfrac{\ln(\cos x)}{-3x^2}+\frac{1}{3}$$ Apply the rule again: $$=\lim_{x\to 0}\dfrac{\sin x}{6x\cos x}+\frac{1}{3}$$ $$=\frac{1}{2}$$

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  • $\begingroup$ It actually surprised me how easily l'Hospital's rule can deal with something like $1-(\cos x)^{\sin x}$. $\endgroup$
    – ryan w.
    Jan 21, 2017 at 18:05
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You can simplify this a bit by using some standard limits, like $$\lim_{x\to0} \frac{e^x-1}{x} = 1$$ and $$\lim_{x\to0} \frac{\ln(1+x)}{x}=1$$. Therefore, $$\lim_{x\to0} \frac{1-(\cos x)^{\sin x}}{x^3} = -\lim_{x\to0}\frac{e^{\sin x\ln(\cos x)}-1}{\sin x\ln(\cos x)}\cdot\frac{\sin x\ln(\cos x)}{x^3} = -\lim_{x\to0} \frac{\sin x\ln(1+\cos x-1)}{\cos x-1}\cdot \frac{\cos x-1}{x^3} = -\lim_{x\to0} \frac{\sin x\cos x-\sin x}{x^3} = -\lim_{x\to0} \frac{\sin(2x)-2\sin x}{2x^3}$$ And the last limit is indeed a lot more approachable than the initial one.

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  • $\begingroup$ MathJax hint: all the functions will come out with proper font ans spacing if you escape them with a backslash just like you did lim. \sin x gives $\sin x$ instead of sin x giving $sin x$ $\endgroup$
    – Ross Millikan
    Jan 21, 2017 at 19:19
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    $\begingroup$ In other words, one rewrites the limit to be computed as the limit of $$\frac{1-(\cos x)^{\sin x}}{x^3} = \frac12\frac{e^{\sin x\cdot\ln(\cos x)}-1}{\sin x\cdot\ln(\cos x)}\cdot\frac{\ln\cos x}{\cos(x)-1}\cdot\left(\frac{\sin(x/2)}{x/2}\right)^2\cdot\frac{\sin x}x$$ and one uses the fact that, when $x\to0$, $$\sin x\cdot\ln(\cos x)\qquad\cos(x)-1\qquad x/2\qquad x$$ all converge to $0$, while, when $t\to0$, the limits of the ratios $$\frac{e^t-1}t\qquad\frac{\ln(1+t)}t\qquad\frac{\sin t}t$$ are all equal to $1$ as derivatives at $t=0$ of the functions $e^t$, $\ln(1+t)$ and $\sin t$. Thus ... $\endgroup$
    – Did
    Jan 22, 2017 at 9:02
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    $\begingroup$ $$\frac{1-(\cos x)^{\sin x}}{x^3}\to\frac12\cdot1\cdot1\cdot1^2\cdot1=\frac12$$ using only the three derivatives at $0$ listed above. Nice. $\endgroup$
    – Did
    Jan 22, 2017 at 9:02
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You can greatly simplify the computations if you remove factors that you know have limit $1$. First, the derivative of $f(x)=(\cos x)^{\sin x}$ is obtained by $\log f(x)=\sin x\log\cos x$, so it is $$ \frac{f'(x)}{f(x)}=\cos x\log\cos x-\frac{\sin^2x}{\cos x}= \cos x(\log\cos x-\tan^2x) $$ so after the first application of l'Hôpital you get $$ \lim_{x\to0}f(x)\cos x\frac{\tan^2x-\log\cos x}{3x^2} $$ and you can disregard $f(x)\cos x$, because it has limit $1$. Therefore you need $$ \lim_{x\to0}\frac{\tan^2x-\log\cos x}{3x^2}= \lim_{x\to0}\frac{2\tan x(1+\tan^2x)+\tan x}{6x}= \lim_{x\to0}\frac{3\tan x+2\tan^3x}{6x}=\frac{1}{2} $$

A different strategy is using Taylor expansions: you can use $$ (\cos x)^{\sin x}=(\cos^2x)^{(\sin x)/2}= (1-\sin^2x)^{\sin x/2} $$ so you can rewrite your limit as $$ \lim_{x\to0}\frac{1-(1-\sin^2x)^{(\sin x)/2}}{\sin^3x}\frac{\sin^3x}{x^3} = \lim_{t\to0}\frac{1-(1-t^2)^{t/2}}{t^3} $$ because the second fraction has limit $1$ and you can do the substitution $t=\sin x$, which is bijective (and continuous) in a neighborhood of $0$.

Now $$ (1-t^2)^{t/2}=\exp\left(\frac{t}{2}\log(1-t^2)\right) = 1+\frac{t}{2}\log(1-t^2)+o(t^3)=1-\frac{t^3}{2}+o(t^3) $$ and so you have $$ \lim_{t\to0}\frac{1-1+t^3/2+o(t^3)}{t^3}=\frac{1}{2} $$

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