Question

How do you find the limit of #(sin(-2x)) / x# as x approaches 0?

Answer 1

I assume you're familiar with the limit

$\displaystyle\lim_{x\to0}\frac{\sin ax}{ax}=1$

for $a\neq0$. We write the given limit in this form:


$\displaystyle\lim_{x\to0}\frac{\sin(-2x)}x=-2\cdot\lim_{x\to0}\frac{\sin(-2x)}{-2x}$

The limit on the RHS is 1, so we're left with -2.