Question

Use the L'Hopital's Rule to find the limit: \[L=\lim_{x \rightarrow 0}\frac{ \sin(2x)-2x }{ 4x^3 }\]

Answer 1

so you know how to apply L'Hopital's rule?

get the derivative of numerator and denominator separately.

let us know what u get

Answer 2

I got this:
\[\lim_{x \rightarrow 0}\frac{ -3\sin(2x)-2xcos(2x)-4x }{ 4x^4 }\]
??

Answer 3

you applied the u/v rule??

thats not necessary :)

take the derivative of numerator and denominator `separately`

derivative of sin 2x - 2x is?

derivative of 4x^3 is ??