Finding $\lim_{x\rightarrow 0}\frac{10^x-2^x-5^x+1}{x\tan(x)}= $? How to calculate the following limit : $$\lim_{x\rightarrow 0}\frac{10^x-2^x-5^x+1}{x\tan(x)}$$
Thanks in advance.

Notice that by the Taylor series we have
$$a^x=\exp(x\ln a)=_0 1+x\ln a+\frac{\ln^2 a}2x^2+o(x^2)$$
and
$$x\tan x= x^2+o(x^2)$$
so we find easily that
$$\lim_{x\to0}\frac{10^x-2^x-5^x+1}{x\tan x}=\frac{\ln^2(10)-\ln^22-\ln^25}{2}$$
