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I am given the following limit to solve, presumably without knowledge of L'Hôpital's rule:

$$\lim_{x\to0}\left({\frac{x}{1-\cos x}}\right)^2$$

I tried using trigonometric identities (namely Pythagorean) to solve it, but with no luck.

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  • $\begingroup$ Figure out $\lim_{x \to 0} \frac{1 - \cos x}{x}$ and go from there. $\endgroup$ – user98602 Dec 16 '15 at 18:46
  • $\begingroup$ Where you mistaking $(1-cos)^2$ with $1-cos^2(x)$? $\endgroup$ – Red Dec 16 '15 at 18:47
  • $\begingroup$ $1-\cos x=2\sin^2 \frac x2$. Try this, the sequence should be divergent i.e. $\lim \dots =\infty$. $\endgroup$ – user249332 Dec 16 '15 at 18:51
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$\displaystyle \lim_{x\to0}\left({\frac{x}{1-\cos x}}\right)^2=$

$\displaystyle = \lim_{x\to0}\left({\frac{x}{2\sin^2 \frac x2}}\right)^2=$

$\displaystyle = \lim_{x\to0}\left({\frac{ \frac{x}{2}}{\sin^2 \frac x2}}\right)^2=$

$\displaystyle = \lim_{x\to0}\left({\frac{ \frac{x}{2}}{\sin \frac x2}} \cdot \frac{1}{\sin \frac x2}\right)^2=$

$\displaystyle =\left(1\cdot \frac{1}{0} \right)^2=$

$=\infty$

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  • $\begingroup$ I see you used the identity $\cos{2x}=1-\sin^{2}{x}$. Did you use the product rule for limits between your third and fourth step? $\endgroup$ – Steve Dec 16 '15 at 20:03
  • $\begingroup$ @Steve yes, I used $ sin^2x = sinx \cdot sinx $ and $ \frac{sinx }{x} \rightarrow 1$ when$ x \rightarrow 0$ $\endgroup$ – klyn Dec 16 '15 at 20:10
  • $\begingroup$ Sorry, I meant in between your fourth and fifth step. You seem to be finding the limit of the products even though the expression is raised to the second power, which I didn't think was possible. $\endgroup$ – Steve Dec 16 '15 at 20:16
  • $\begingroup$ @Steve yes, you can think as $\displaystyle \lim_{x\to0}{\frac{ \frac{x}{2}}{\sin \frac x2}} \cdot \frac{1}{\sin \frac x2} \cdot \frac{ \frac{x}{2}}{\sin \frac x2} \cdot \frac{1}{\sin \frac x2}=1\cdot \frac{1}{0}\cdot1\cdot \frac{1}{0} = 1 \cdot \infty \cdot 1 \cdot \infty = \infty$ $\endgroup$ – klyn Dec 16 '15 at 20:31
  • $\begingroup$ But now that the expression is written that way, $\lim _{ x\rightarrow 0 }{ \frac { 1 }{ sin\frac { x }{ 2 } } }$ is undefined since $\lim _{ x\rightarrow { 0 }^{ - } }{ \frac { 1 }{ sin\frac { x }{ 2 } } } =-\infty$ and $\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { 1 }{ sin\frac { x }{ 2 } } } =\infty$ $\endgroup$ – Steve Dec 16 '15 at 20:42
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$$\frac x{1-\cos x}=\frac x{1-\cos x}\cdot\frac{1+\cos x}{1+\cos x}=\frac{x(1+\cos x)}{\sin^2x}$$

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