Trigonometric limit solution

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 $f(x)= \frac{2}{\left( \frac{\sin x}{x}\right)^2}$. – André Nicolas May 17 '12 at 5:05

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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But how do I convert my expression to that form? sin²x is dividing in this case – Jorge Zapata May 17 '12 at 4:49
@JorgeZapata, only squating and inverting. – Gastón Burrull May 17 '12 at 4:50
@Jorge: Not if you pick the right $f$. HINT: $f(x)$ is not $\sin^2 x$. – Brian M. Scott May 17 '12 at 4:58
@JorgeZapata If you know a value of some limit, in particular you know that limit exists and almost all aritmethic properties of limits you can use without problems. The properties that may you use are squaring and inverting. Use $f(x)=\frac{\sin x}{x}$ because is the only limit that you known but it is enought. – Gastón Burrull May 17 '12 at 5:07
@Jorge: $$\lim_{x\to 0}\frac{2x^2}{\sin^2 x}=2\lim_{x\to 0}\frac1{\frac{\sin^2 x}{x^2}}=\dots\;?$$ – Brian M. Scott May 17 '12 at 5:08

$2\cdot \lim_{x\to 0} \dfrac{1}{\frac{\sin x}{x}} \cdot\dfrac{1}{\frac{\sin x}{x}} = 2(\frac{1}{1} \cdot \frac{1}{1}) = 2$

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