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In my assignment I have to solve the following question. I know the answer, but I keep getting it wrong, and I don't know how to solve it.

$$\lim_{x \to 0} \frac{1-\cos x}{x\sin x}$$

I have tried several things, but first I tried to multiply by $(1+\cos x)$, both numerator and denumerator, to get $$\lim_{x \to 0} \frac{1-\cos^2x}{(x\sin x)(1+\cos x)}$$ I keep getting that the limit is equal to $0$, by calculating (for example) $$\lim_{x \to 0} \frac{\frac{1}{\cos^2x}-\frac{\cos^2x}{\cos^2x}}{\frac{(x\sin x)}{\cos^2x}\frac{(1+\cos x)}{\cos^2x}},$$

which equals $$\frac{1-1}{1-1}$$

However, I know the answer is $\frac{1}{2}$.

Any ideas?

Thanks

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  • $\begingroup$ Can you use L'Hôspital? $\endgroup$ – Vinícius Novelli May 11 '15 at 16:04
  • $\begingroup$ @ViníciusNovelli no, sorry, I didn't study it. $\endgroup$ – Alan May 11 '15 at 16:05
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    $\begingroup$ How are you getting that that limit is equal to zero? $\endgroup$ – Bob Krueger May 11 '15 at 16:05
  • $\begingroup$ You're correct up until you start trying to manipulate the $1-\cos^2x$. You should start there and make use of the identity $\sin^2x+\cos^2x=1$. $\endgroup$ – abiessu May 11 '15 at 16:12
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$$ \frac{1-\cos^2{x}}{x\sin x(1+\cos x)}=\frac{\sin^2{x}}{x\sin x(1+\cos x)}=\frac{\sin x}{x(1+\cos x)}=\big(\frac{\sin x}{x}\big)\frac{1}{1+\cos x} $$

Does that look familiar now?

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  • $\begingroup$ @Thank you! I know that $$\frac{sinx}{x}$$ has a limit of 1, and that $1+cosx$ also has a limit of 1. $\endgroup$ – Alan May 11 '15 at 16:11
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    $\begingroup$ $1+\cos x$ has a limit of 2, right? $\cos 0 = 1$. $\endgroup$ – Vinícius Novelli May 11 '15 at 16:12
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Multiplying by $\displaystyle \frac{\sin x}{x}$ immediately does the job, recalling the first two fundamental trigonometric limits.

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If you know that $$\begin{align}\lim\limits_{x\to0} \frac{\sin x}x&=1 \tag{1}\\ \lim\limits_{x\to0} \frac{1-\cos x}{x^2}&=\frac12 \tag{2} \end{align}$$ then it is not difficult to combine them to get: $$\lim\limits_{x\to0} \frac{1-\cos x}{x\sin x} = \lim\limits_{x\to0} \frac{1-\cos x}{x^2} \cdot \frac{x}{\sin x} = \frac 12 \cdot 1 = \frac12.$$

The limits $(1)$ and $(2)$ are quite well-known. The first one can be found, for example, here. And the second one can be found, for example, here or here.

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hint:Use these facts:

$$1-\cos x = 2\sin^2\left(\frac{x}{2}\right), \sin x = 2\sin(\frac{x}{2})\cos (\frac{x}{2}),\displaystyle \lim_{x\to 0} \dfrac{\sin (\frac{x}{2})}{\frac{x}{2}}=1$$

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In order to be comprehensive I add an answer based on Taylor expansions

$$1-\cos{x}=\frac{x^2}{2}+o(x^2)$$ $$x\sin{x}=x^2+o(x^2)$$

This yields

$$\frac{1-\cos{x}}{x\sin{x}}=\frac{1}{2}+o(1)$$

And the limit is $\frac{1}{2}$

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You can solve this trigonometric limit
also with the l'Hopital's rule: $$\lim_{x \to 0} \frac{1-\cos x}{x\sin x}=$$ $$\lim_{x \to 0} \frac{\sin x}{\sin x+\cos x}=$$ $$\lim_{x \to 0} \frac{\cos x}{\cos x+\cos x-x\cdot \sin x}=\frac {1}{2}$$

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