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I am trying to find the limit of $\lim_{x \downarrow 0} (\frac{1}{sinx}- \frac{1}{x})^x$.

My current progress:

$\lim_{x \downarrow 0} (\frac{1}{\sin x}- \frac{1}{x})^x = \lim_{x \downarrow 0} (\frac{x-\sin x}{x\sin x})^x = \lim_{x \downarrow 0} e^{xln(\frac{x-\sin x}{x \sin x})} = e^{\lim_{x \downarrow 0}xln(\frac{x-\sin x}{x \sin x})} = e^{\lim_{x \downarrow 0}\frac{ln(\frac{x-\sin x}{x \sin x})}{1/x}}$

This is where I used l'Hospital, but after 2 iterations of l'Hospital it doesn't look like I'll get anywhere with it.

Expression after first l'Hospital: (I'll not write the e as the base, so that the limit is easier to read)

$\lim_{x \downarrow 0}\frac{-x^3\sin x\cos x - x \sin^3x}{x\sin^2x - \sin^3x} = \lim_{x \downarrow 0}\frac{-x^3\cos x - x \sin^2x}{x\sin x - \sin^2x}$

Am I missing something or is it just tedious to find the limit of this expression?

After the second iteration of l'Hospital, I've got this term:

$\lim_{x \downarrow 0}\frac{-3x^2\cos x + x^3\sin x - \sin^2x -2x\sin x\cos x}{x\cos x + \sin x -2\sin x\cos x} = "\frac{0}{0}"$

is that term correct? If so, I'll try to apply l'Hospital a third time.

Meik Vtune
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3 Answers3

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you can write your term in the form $$\frac{-x^3\cos(x)-x\sin(x)^2}{x\sin(x)-\sin(x)^2}$$ and apply L'Hospitel one more time

Dr. Sonnhard Graubner
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$$\begin{align} x\log\left(x-\sin x\over x\sin x\right) &=x\log\left(x-(x-{1\over6}x^3+{1\over120}x^5-\cdots)\over x^2-{1\over6}x^4+{1\over120}x^6-\cdots\right)\\ &=x\log\left({1\over6}x-{1\over120}x^3+\cdots\over1-{1\over6}x^2+{1\over120}x^4-\cdots \right)\\ &=x\log x+x\log\left({1\over6}-{1\over120}x^2+\cdots\right)-x\log\left(1-{1\over6}x^2+{1\over120}x^4-\cdots\right) \end{align}$$

It's easy (via L'Hopital, for example) to show that $x\log x\to0$ as $x\to0^+$, while the other two expressions tend to $0\log{1\over6}=0$ and $0\log1=0$. Thus

$$\left({1\over\sin x}-{1\over x}\right)^x\to e^0=1\quad\text{as}\quad x\to0^+$$

Barry Cipra
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  • 2 Questions: 1) where did you get $x log x$ after your third equal? 2) can you just factor out the x in the log? – Meik Vtune Jun 08 '17 at 18:12
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    @MeikVtune, I'm using the logarithmic identity $\log(ab)=\log a+\log b$ with $ab=x({1\over6}-{1\over120}x^2+\cdots)$ – Barry Cipra Jun 08 '17 at 18:20
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If $ \lim\ (\frac{1}{\sin\ x} - \frac{1}{x} )^x=L$, then $$ \lim\ x\cdot \ln\ (\frac{1}{\sin\ x} - \frac{1}{x} )=\ln\ L$$

Here by Taylor series, $$ x \ln\ ( \frac{1}{\sin\ x} - \frac{1}{x} )= x \ln\ \frac{x-\sin\ x}{x\sin\ x} =x\ln\ \frac{x/6 + O(x^2)}{1 + O(x^2)} =\frac{\ln\ (\frac{x}{6} + O(x^2) ) }{1/x}$$

L'Hospital implies that $$ \ln\ L = \lim\ \frac{ \frac{ 1/6 + O(x) }{ x/6 + O(x^2) } }{ -1/x^2} =\lim\ - \frac{ 1/6 + O(x) }{ 1/6 + O(x) } x =0 $$

Hence $L=1$

HK Lee
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