TITLE: Determine the limit using Taylor expansions:
QUESTION [5 upvotes]: I struggle with this one—maybe someone could point me in the right direction.
$$\lim_{x\to 0} \frac{5^{(1+\tan^2x)} -5}{1-\cos^2x}$$
Getting the Taylor series expansion for $\tan^2x$ and $\sin^2x$ is no problem, but I struggle with getting further along at this step:
$$\frac{5^1 \cdot 5^{x^2} \cdot 5^{(2/3)x^4} -5}{x^2 - \frac{x^4}3}$$
Any help would be greatly appreciated!

REPLY [2 votes]: Using $\exp(x)=1+x+O(x^2)$ and $\tan^{2}(x) = x^{2}+O(x^{4})$
\begin{align*}5^{\tan^{2}(x)} &=\exp(\ln(5)\tan^{2}(x))\\
&=\exp(\ln(5)(x^2+O(x^4)))\\
&=1+\ln(5)(x^2+O(x^{4}))+ O([x^2+O(x^{4})]^2)\\
&=1+\ln(5)x^{2}+O(x^{4})
\end{align*}
Therefore, using $1-\cos^{2}(x)=x^{2}+O(x^{4})$:
$$\lim_{x\to 0}\frac{5(5^{\tan^{2}(x)}-1)}{1-\cos^{2}(x)}=\lim_{x\to 0}5\frac{1+\ln(5)x^2+O(x^4)-1}{x^{2}+O(x^{4})}$$
I am sure you can take it from here