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.

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.
