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)(\sin ax)/(ax)=1`

for
`a\\eq0`. We write the given limit in this form:



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

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