Find : $\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right)$ with l'Hôpital's rules I want to calculate :
$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right)$$
with l'Hôpital's rules.
I get
$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right) = \lim\limits_{x \rightarrow 0}\left(\frac{x-\sin(x)}{x\sin(x)}\right) = \lim\limits_{x \rightarrow 0}\left(\frac{1-\cos(x)}{\sin(x)+x\cos(x)}\right)$$
but I don't see useful next steps.
 A: Since the question has been tagged as "real analysis", I assume that the OP is familiar with Taylor series: The idea of using L'hopital's rule for trigonometric functions is generally not good because when you differentiate a trigonometric function, you again get a new trigonometric function. + differentiation of the multiplication of two terms (in the denominator) makes it more complicated. So, L'hopital is not very efficient in this problem.
You can do it more easily using approximates of the Taylor series for $\sin(x)$:
$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right)=\lim\limits_{x\to 0}\frac{x-\sin(x)}{x\sin(x)}=\lim\limits_{x\to 0}\frac{x^3/6}{x\sin(x)}=\lim\limits_{x\to 0}x/6\frac{x}{\sin(x)}=0\times1=0$$
Here's an alternative way using only L'hopital's rule:
$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right)=\lim\limits_{x\to 0}\frac{x-\sin(x)}{x\sin(x)}=\lim\limits_{x\to 0}\frac{1-\cos(x)}{\sin(x)+x\cos(x)}=\lim\limits_{x\to 0}\frac{\frac{1-\cos(x)}{x}}{\frac{\sin(x)}{x}+\cos(x)}=\lim\limits_{x\to 0}\frac{1-\cos(x)}{x}\times \lim\limits_{x\to 0}\left(\frac{1}{\frac{\sin(x)}{x}+\cos(x)}\right)=0\times \frac{1}{2}=0$$
Use L'hopitals rule separately to prove that:
$$\lim\limits_{x\to 0}\frac{\sin(x)}{x}=1$$
$$\lim\limits_{x\to 0}\frac{1-\cos(x)}{x}=0$$
A: Not taking derivative again, we can also look that $1-\cos x = 2 \sin^2 (x/2)$. Using this you get zero as limit.
$$\lim\limits_{x \rightarrow 0}\left(\frac{\tfrac{2\sin^2(x/2)}{x}}{\tfrac{\sin(x)}{x}+\cos(x)}\right) = 0$$
A: You can also use Taylor series:
$$\sin(x)=x-\frac{x^3}{6}+O(x^4)$$
So your function:
$$\frac{1}{\sin(x)}-\frac{1}{x}=\frac{x-\sin(x)}{x\sin(x)}=\frac{\frac{x^3}{6}+O(x^4)}{x^2-\frac{x^4}{6}+O(x^5)}=\frac{\frac{x}{6}+O(x^2)}{1-\frac{x^2}{6}+O(x^3)} \to \frac{0}{1}=0$$
A: just use it again
$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right) = \lim\limits_{x \rightarrow 0}\left(\frac{x-\sin(x)}{x\sin(x)}\right) =\\ \lim\limits_{x \rightarrow 0}\left(\frac{1-\cos(x)}{\sin(x)+x\cos(x)}\right) =
\lim\limits_{x \rightarrow 0}\left(\frac{\sin(x)}{\cos(x)-x\sin(x)+\cos(x)}\right) = \lim\limits_{x \rightarrow 0}\left(\frac{0}{1-0\cdot0+1}\right)=0$$
A: As it turns out, you just have to use L'Hospital's rule twice and observe that on the second try the denominator is no longer zero when you use directly substitution:
\begin{align}
\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin x} - \frac{1}{x}\right)
&=\lim\limits_{x \rightarrow 0}\frac{x -\sin x}{x\sin x}\\
&=\lim\limits_{x \rightarrow 0}\frac{(x -\sin x)'}{(x\sin x)'}\\
&=\lim\limits_{x \rightarrow 0}\frac{x' -(\sin x)'}{x'\sin x+x(\sin x)'}\\
&=\lim\limits_{x \rightarrow 0}\frac{1 - \cos x}{\sin x+x \cos x}\\
&=\lim\limits_{x \rightarrow 0}\frac{(1 - \cos x)'}{(\sin x+x \cos x)'}\\
&=\lim\limits_{x \rightarrow 0}\frac{1' - (\cos x)'}{(\sin x)'+x' \cos x +x(\cos x)'}\\
&=\lim\limits_{x \rightarrow 0}\frac{\sin x}{\cos x + \cos x - x \sin x}\\
&=\lim\limits_{x \rightarrow 0}\frac{\sin x}{2\cos x - x \sin x}\\
&=\frac{0}{2\cdot 1 - 0 \cdot 0}\\
&=\frac{0}{2}\\
&=0
\end{align}
A: If you are forced to use the infamous L'Hospital's Rule then you should first try to simplify the expression and then apply L'Hospital's Rule.
We can proceed as follows
\begin{align}
L&=\lim_{x\to 0}\frac{1}{\sin x} - \frac{1}{x}\notag\\
&=\lim_{x\to 0}\frac{x-\sin x} {x\sin x} \notag\\
&=\lim_{x\to 0}\frac{x-\sin x} {x^{2}}\cdot\frac{x}{\sin x} \notag\\
&=\lim_{x\to 0}\frac{x-\sin x} {x^{2}} \notag \\
&=\lim_{x\to 0}\frac{1-\cos x} {2x}\text{ (via L'Hospital's Rule)} \notag \\
&=\frac{1}{2}\lim_{x\to 0}\frac{1-\cos^{2}x}{x(1+\cos x)} \notag\\
&=\frac{1}{4}\lim_{x\to 0}x\cdot\left(\frac{\sin x} {x} \right) ^{2}\notag\\
&=\frac{1}{4}\cdot 0\cdot 1^{2}\notag \\
&=0\notag
\end{align} 
A: Sum the fraction Then apply easily L'Hopiyal rule twice
$$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)= \lim_{x \rightarrow 0}\left(\frac{\sin x - x}{x\sin x}\right) \\= \lim_{x \rightarrow 0}\left(\frac{\cos x - 1}{\sin x +x\cos x}\right)=\lim_{x \rightarrow 0}\left(\frac{-\sin x }{2\cos x -x\sin x }\right)=0$$
