# 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:

1. $e^x = 1 + x + \frac{1}{2} x^2 + \frac{1}{6} x^3 + o(x^3)$
2. $\sin x = x - \frac{1}{6} x^3 + o(x^4)$
3. $e^{\sin x} = 1 + x - \frac{1}{6} x^3 + o(x^4)$
4. $\sin 2x = 2x + o(x^2)$
5. $\sin^3 2x = 8x^3 + o(x^6)$

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?

• The first equation is wrong. correct one is $\mathbb{e}^x = 1 + x + \frac{1}{2} x^2 + \frac{1}{\color{red}{6}} x^3 + o(x^3)$ – Shane Chern Jan 28 '13 at 12:46
• Argh, what a silly mistake! Fixed that and an another copy&paste error. – hey hey Jan 28 '13 at 12:50
• You forgot to add $\frac{1}{2}\sin^2 x=\frac{1}{2}x^2+o(x^3)$ and $\frac{1}{6}\sin^3 x=\frac{1}{6}x^3+o(x^3)$... – Shane Chern Jan 28 '13 at 12:59

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