Calculate limit of large function How can I calculate this:
$$\lim_{x\to 0}{\frac{(\sin x)^{13} - \ln(1 + (\sin x) ^{13} )} {(\tan x) ^{26} }} $$
I tried taking $\frac{1}{{\tan^{13} x}} $ common and then evaluating the limit and I got answer zero but the answer is $\frac{1}{2}$. Any suggestions?
Let $\sin^{13}x=u$

$$\lim_{u\to0}\dfrac{u-\ln(1+u)}{u^2}\cdot\lim_{x\to0}\cos^{26}x =\lim_{u\to0}\dfrac{u-\ln(1+u)}{u^2}$$

Now use Series Expansion or L'Hospital's Rule
I think you need to use approximations ie $$\tan(x)=\sin(x)\approx x $$ for small x and taylor expansion for $ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}...$