$\lim_{x \to0} (\sin(2x)-2x)/x^3$

Question

Evaluate $$ \lim_{x\to0} \frac{\sin(2x)-2x}{x^3} $$

I thought that I could use L'Hopital's rule to get to an answer of $-1$ but according to the answer manual that isn't correct. What am I doing wrong?

Answer 1

This is the solution, you keep using L Hospital's Rule until you get the denominator as a number or an integer instead of x, since x-> 0 means the denominator will be 0 which wont solve our problem.

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Answer 2

A better method would be using Taylor series: $$ \lim_{x\to 0} \frac{\sin(2x)-2x}{x^3} = \lim_{x\to 0} \frac{2x-\frac{(2x)^3}{3!}+O(x^5)-2x}{x^3} = -\frac{4}{3}, $$ where we used $\sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$.