Determine the limit using Taylor expansions: 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!
 A: $\lim\limits_{x\to0}\dfrac{5^{1+\tan^2(x)}-5}{1-\cos^2(x)}=$
$=5\lim\limits_{x\to0}\dfrac{5^{\tan^2(x)}-1}{\sin^2(x)}=$
$=5\lim\limits_{x\to0}\left[\dfrac{5^{\tan^2(x)}-1}{\tan^2(x)}\cdot\dfrac1{\cos^2(x)}\right]=$
$=5\lim\limits_{x\to0}\dfrac{5^{\tan^2(x)}-1}{\tan^2(x)}\cdot\lim\limits_{x\to0}\dfrac1{\cos^2(x)}\underset{\overbrace{\text{by letting }y=\tan^2(x)}}{=}$
$=5\lim\limits_{y\to0}\dfrac{5^y-1}y\cdot\lim\limits_{x\to0}\dfrac1{\cos^2(x)}=$
$=5\cdot\ln5\cdot1=$
$=5\ln5\;.$
Addendum :
If you want to calculate the limit by using Taylor expansions, you can proceed in the following way :
$\dfrac{5^{1+\tan^2(x)}-5}{1-\cos^2(x)}=$
$=5\cdot\dfrac{5^{\tan^2(x)}-1}{\sin^2(x)}=$
$=5\cdot\dfrac{5^{\tan^2(x)}-1}{\tan^2(x)}\cdot\dfrac1{\cos^2(x)}\underset{\overbrace{\text{by letting }y=\tan^2(x)}}{=}$
$=5\cdot\dfrac{5^y-1}y\cdot\dfrac1{\cos^2(x)}=$
$=5\ln5\cdot\dfrac{e^{y\ln5}-1}{y\ln5}\cdot\dfrac1{\cos^2(x)}=$
$=5\ln5\cdot\dfrac{y\ln5+\frac{(y\ln5)^2}2+O(y^3)}{y\ln5}\cdot\dfrac1{\left(1-\frac{x^2}2+O(x^4)\right)^2}=$
$=5\ln5\cdot\dfrac{1+\frac{y\ln5}2+O(y^2)}{1-x^2+O(x^4)}\xrightarrow{\color{blue}{x\to0}}5\ln5\;.$
It is not necessary to write only Taylor expansions in powers of $x$, we can also write a Taylor expansion in powers of $y$ which is an infinitesimal function of $x$ as $x\to0\;.$
A: 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
A: $$\lim_{x\to 0} 5 \ln(5) + 5/2 x^2 \ln (5) (2 + \ln (5)) + O(x^4)= 5 \ln(5)$$
https://www.wolframalpha.com/input/?i=taylor+series+%5Cfrac%7B5%5E%7B%281%2B%5Ctan%5E2%28x%29%29%7D+-5%7D%7B1-%5Ccos%5E2%28x%29%7D
