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Why does the following limit equals 2:

$\lim_{x \to 0}\frac{2x^2}{\sin^2 x}=2$

I can't find a trigonometric conversion to get that result.

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    Let $f(x)$ be your function. Then $f(x)= \frac{2}{\left( \frac{\sin x}{x}\right)^2}$.2012-05-17

3 Answers 3

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Use the basic limit $\lim_{x\to 0}\frac{\sin x}x=1$ to derive this one. Then use some basic properties of limits. For example,

$\lim_{x\to a}\frac1{f(x)}=\frac1{\lim\limits_{x\to a}f(x)}\;,$

if the limit in the denominator is not zero.

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    Thanks @BrianM.Scott my problem was figure it out how to get that denominator. $2\lim_{x \to0} \frac{x^2}{sin^2x}*\frac{\frac{1}{x^2}}{\frac{1}{x^2}}$2012-05-17
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$2\cdot \lim_{x\to 0} \dfrac{1}{\frac{\sin x}{x}} \cdot\dfrac{1}{\frac{\sin x}{x}} = 2(\frac{1}{1} \cdot \frac{1}{1}) = 2$

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    For some basic information about writing math at this site see e.g. [here](http://meta.math.stackexchange.com/questions/5020/), [here](http://meta.stackexchange.com/a/70559/155238), [here](http://meta.math.stackexchange.com/questions/1773/) and [here](http://math.stackexchange.com/editing-help#latex).2013-01-31
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use Bernoulli's rule , wiki :http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule