# Solve this limit without using L'Hôpital's rule

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.

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

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

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

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

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

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

$=\infty$

• 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? – Steve Dec 16 '15 at 20:03
• @Steve yes, I used $sin^2x = sinx \cdot sinx$ and $\frac{sinx }{x} \rightarrow 1$ when$x \rightarrow 0$ – klyn Dec 16 '15 at 20:10
• 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. – Steve Dec 16 '15 at 20:16
• @Steve yes, you can think as $\displaystyle \lim_{x\to0}{\frac{ \frac{x}{2}}{\sin \frac x2}} \cdot \frac{1}{\sin \frac x2} \cdot \frac{ \frac{x}{2}}{\sin \frac x2} \cdot \frac{1}{\sin \frac x2}=1\cdot \frac{1}{0}\cdot1\cdot \frac{1}{0} = 1 \cdot \infty \cdot 1 \cdot \infty = \infty$ – klyn Dec 16 '15 at 20:31
• 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 $\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { 1 }{ sin\frac { x }{ 2 } } } =\infty$ – Steve Dec 16 '15 at 20:42

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