Calculate $\lim_{x\rightarrow0}\frac{(e^{\sin x}+ \sin x)^{\frac{1}{\sin x}}-(e^{\tan x}+ \tan x)^{\frac{1}{\tan x}}}{x^3}$ How to calculate the following limit?
$$\lim_{x\rightarrow0}\frac{(e^{\sin x}+ \sin x)^{\frac{1}{\sin x}}-(e^{\tan x}+ \tan x)^{\frac{1}{\tan x}}}{x^3}$$
I thought of L'Hopital's rule, Taylor expansion, and limit the form of $e^x$, but the presence of $\sin x$ and $\tan x$ make it hard to apply them. Could anyone give me a hint?
 A: Hint 1: Consider the Taylor series of $f(\sin(x)) - f(\tan(x))$ for a general function $f$. What information do you need about $(e^x + x)^{1/x}$ to solve the problem?
Hint 2:

 The information can be gotten from the Taylor series for $\ln(e^x + x)/x$, which is easier to find.

Full solution:

 You can verify by substitution of Taylor series that if we have a function $f(x) = \sum_{i=0}^\infty a_i x^i$, then $f(\sin x) - f(\tan x) = -f'(0) x^3/2 + O(x^4)$. So we simply need to find the derivative of $f(x) = (e^x + x)^{1/x}$ at $0$. Since we only need the first derivative, it'll be easier to look at $\ln(f(x)) = \ln(e^x + x)/x$ instead. The numerator can be easily expanded in power series, and we get $$\frac{\ln(e^x + x)}{x} = \frac{1}{x}\left(2x - \frac{3}{2} x^2 + \frac{11}{6} x^3 + O(x^4)\right) = 2 - \frac{3}{2}x + \frac{11}{6}x^2 + O(x^3)$$ So we have $\ln(f(0)) = 2$ and $\ln(f(x))'|_0 = -3/2$. Using $\ln(f(x))' = f'(x)/f(x)$, we have $f'(0) = -3e^2/2$. Thus, \begin{multline}\lim_{x\rightarrow 0} \frac{(e^{\sin x} - \sin x)^{1/\sin x} - (e^{\tan x} - \tan x)^{1/\tan x}}{x^3} \\= \lim_{x\rightarrow 0}\frac{1}{x^3}\left[\left(-\frac{1}{2}\right)\left(-\frac{3e^2}{2}\right)x^3 + O(x^4)\right) = \frac{3e^2}{4} \end{multline}

A: Hint: $$(e^{\sin x}+\sin x)^{\frac {1}{\sin x}}=e^{\frac {\ln(e^{\sin x}+\sin x)}{\sin x}}=e^{1+\frac {\ln(1+\frac {\sin x}{e^{\sin x}})}{\sin x}}=e(e^{e^{-\sin x}})=e^{2-\sin x+\frac {\sin^2 x}{2}}$$
Now, one more application of Taylor Series would create a polynomial in $\sin x$. To do this make sure to keep $e^2$ separate first. You can do the same thing with $\tan x$.
So, you get a polynomial in $\sin x$ and $\tan x$, after which you can evaluate the limit using L'Hôspital rule, because the indeterminate part of it will factorise to create a simple limit.
