Question

limit x to 0 of sin(4x)/2x

Answer 1

Are you allowed to use L'Hopital's rule? If so, it's a very simple limit.

Answer 2

i got it

Answer 3

\[\LARGE \lim_{x\to 0} \frac{\sin 4x}{2x} \]

Multiply numerator and denominator by \[\LARGE \frac44\]

you'll have..

\[\LARGE \lim_{x\to 0} \frac{4\cdot \sin 4x}{4\cdot 2x} \]

\[\LARGE \lim_{x\to 0} \frac{\sin 4x}{4x} \cdot \frac{4}{2}\]

You're supposed to know this rule:

\[\LARGE \lim_{x\to 0}\frac{\sin x}{x}=1\]

so the answer should be 2