Find $\lim_{x \rightarrow 0} \frac{e^{\sin x} - e^x}{\sin^3 2x}$ I have to find $\displaystyle\lim_{x \rightarrow 0} \frac{e^{\sin x} - e^x}{\sin^3 2x}$ using Taylor polynomials.
Here's what I've done so far:


*

*$e^x = 1 + x + \frac{1}{2} x^2 + \frac{1}{6} x^3 + o(x^3)$

*$\sin x = x - \frac{1}{6} x^3 + o(x^4)$

*$e^{\sin x} = 1 + x - \frac{1}{6} x^3 + o(x^4)$

*$\sin 2x = 2x + o(x^2)$

*$\sin^3 2x = 8x^3 + o(x^6)$


Therefore I can rewrite my limit as: $\displaystyle\lim_{x \rightarrow 0} \frac{-\frac{1}{3} x^3 - \frac{1}{2} x^2 + o(x^3)}{8x^3 + o(x^6)}$
In this form, the limit appers to be $-\frac{1}{24}$, but the correct result is $-\frac{1}{48}$.
Could you tell me what I'm doing wrong?

Let's rewrite the limit as
$$\displaystyle\lim_{x \rightarrow 0} e^x\times\lim_{x \rightarrow 0}\frac{e^{\sin x-x} - 1}{\sin x-x}\times\lim_{x \rightarrow 0}\left(\frac{2x}{\sin 2x}\right)^3\times \lim_{x \rightarrow 0}\frac{\sin x-x}{8x^3}=\lim_{x \rightarrow 0}\frac{x-x^3/6+O(x^5)-x}{8x^3}=-\frac{1}{48}$$
Q.E.D.
