find the limit of $ (1/\sin\ (x)-1/x)^x$ 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.
 A: 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
A: $$\begin{align}
x\log\left(x-\sin x\over x\sin x\right)
&=x\log\left(x-(x-{1\over6}x^3+{1\over120}x^5-\cdots)\over x^2-{1\over6}x^4+{1\over120}x^6-\cdots\right)\\
&=x\log\left({1\over6}x-{1\over120}x^3+\cdots\over1-{1\over6}x^2+{1\over120}x^4-\cdots \right)\\
&=x\log x+x\log\left({1\over6}-{1\over120}x^2+\cdots\right)-x\log\left(1-{1\over6}x^2+{1\over120}x^4-\cdots\right)
\end{align}$$
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
$$\left({1\over\sin x}-{1\over x}\right)^x\to e^0=1\quad\text{as}\quad x\to0^+$$
A: 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 $L=1$
