Homework Problem, Power Series Limit I am looking to find the solution for:
$$\lim_{x\rightarrow 0} {5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$$
A hint was provided:
transform: ${5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$ to $5y{5^{y-1} - 1 \over y-1}, y = {1\over cos^s(x)} \rightarrow 1$
The transformation is straigt forward:
$tan^2(x) + 1 = {1\over cos^2(x)}  = y \text{ and } \\1-cos^2(x) = ycos^2(x) - cos^2(x) = cos^2(x)(y-1)$ combined we have:
$$5y{(5^{y-1} - 1) \over y-1}$$
as $y \rightarrow 1$ $5y{(5^{y-1} - 1) \over y-1}$ is not defined. Since both, nominator and denominator are $0$ I tried L'Hopital but ended at:
$5 5^{y-1} + 5y(y-1)5^{y-2}-5$ with  $\lim_{y \rightarrow 1} = 5 + 0 - 5 = 0$ and $(y-1)' = 1$
Here I have to stop with no solution. I have also tried to use the quotient rule to differentiate the expression which did not get me anywhere. 

Hint:
$$5\cdot\dfrac{5^{\tan^2x}-1}{\tan^2x}\cdot\dfrac{\tan^2x}{1-\cos^2x}=5\cdot\dfrac{5^{\tan^2x}-1}{\tan^2x}\cdot\dfrac1{\cos^2x}$$
Now as $a=e^{\ln a}$ for $a>0$
$$\lim_{h\to0}\dfrac{a^h-1}h=\cdots=\ln a$$
