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.
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.
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.
$2\cdot \lim_{x\to 0} \dfrac{1}{\frac{\sin x}{x}} \cdot\dfrac{1}{\frac{\sin x}{x}} = 2(\frac{1}{1} \cdot \frac{1}{1}) = 2$
use Bernoulli's rule , wiki :http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule