$\lim_{x \to0} (\sin(2x)-2x)/x^3$
By Emily Wilson •
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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?
$\endgroup$ 41 Answer
$\begingroup$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.
Kindly message me if you have problems or concerns in understanding this.
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