Limiting question:$\displaystyle \lim_{x\to\ 0} \frac{a^{\tan\ x} - a^{\sin\ x}}{\tan\ x -\sin\ x}$ How do I find the value of $$\lim_{x\to\ 0} \frac{a^{\tan\ x} - a^{\sin\ x}}{\tan\ x - \sin\ x}$$
in easy way.
 A: Observe that
$$ \lim_{x \to 0} \frac{\tan x - \sin x}{x^3/2} = 1
$$
Hence
$$\lim_{x\to\ 0} \frac{a^{\tan x} - a^{\sin x}}{\tan x - \sin x} = \lim_{x\to\ 0} \frac{a^{\tan x} - a^{\sin x}}{\tan x - \sin x}\frac{\tan x - \sin x}{x^3/2} = 2\lim_{x \to 0} \frac{a^{\tan x} - a^{\sin x}}{x^3}
$$
Now l'Hopital only need to be applied $3$ times.
A: As Henry W; commented, Taylor series make thigs quite simple.
$$A=a \tan(x) \implies \log(A)=\tan(x)\log(a)=\Big(x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^6\right)\Big)\log(a)$$ $$A=e^{\log(a)}\implies A=1+x \log (a)+\frac{1}{2} x^2 \log ^2(a)+\frac{1}{6} x^3 \left(\log ^3(a)+2 \log
   (a)\right)+O\left(x^4\right)$$ Soing the same with the second term of numerator $$B=a \sin(x) \implies B=1+x \log (a)+\frac{1}{2} x^2 \log ^2(a)+\frac{1}{6} x^3 \left(\log ^3(a)-\log
   (a)\right)+O\left(x^4\right)$$ So the numerator is $$\frac{1}{2} x^3 \log (a)+O\left(x^4\right)$$ Using again the  series for $\sin(x)$ and $\tan(x)$, the denominator is $$\frac{x^3}{2}+O\left(x^4\right)$$ which makes the ratio $$\frac{a^{\tan\ x} - a^{\sin\ x}}{\tan\ x -\sin\ x}=\log (a)+O\left(x^1\right)$$ Using more terms would lead to $$\frac{a^{\tan\ x} - a^{\sin\ x}}{\tan\ x -\sin\ x}=\log (a)+x \log ^2(a)+O\left(x^2\right)$$ showing the limit and how it is approached.
A: Let $u=\tan(x)-\sin(x)$. Note that as $x\to 0$, $u\to 0$.  Then, we can write
$$\begin{align}
\lim_{x\to 0}\frac{a^{\tan(x)}-a^{\sin(x)}}{\tan(x)-\sin(x)}&=\left(\lim_{x\to 0}\frac{a^{\tan(x)-\sin(x)}-1}{\tan(x)-\sin(x)}\right)\,\left(\lim_{x\to 0}a^{\sin(x)}\right)\\\\
&=\left(\lim_{u\to 0}\frac{a^u-1}{u}\right)\,\left(\lim_{x\to 0}a^{\sin(x)}\right)\\\\
&=\log(a)
\end{align}$$
since 
$$\lim_{u\to 0}\frac{a^u-1}{u}=\left.\frac{da^u}{du}\right|_{u=0}=\log(a)$$
