Homework Problem, Power Series Limit

Question

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.

Answer 1

Hint:

If you are using L'Hopital and differentiating with respect to $y$, then \begin{align} \frac{\mathrm{d}}{\mathrm{d}y\,}5^{y-1} &= \frac{\mathrm{d}}{\mathrm{d}y\,}e^{(y-1) \ln5}\\ &=\ln 5 \cdot e^{(y-1) \ln5} \\ &=\ln 5 \cdot 5^{y-1} \end{align}

Answer 2

As $x\to 0$, $$1 - \cos^2(x) = \sin^2(x) \sim x^2$$ Use this to simplify thigs.

Answer 3

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 $a>0$

$$\lim_{h\to0}\dfrac{a^h-1}h=\cdots=\ln a$$

Answer 4

I’m not sure you’re doing the differentiation right. We have

$$\lim_{y\to 1} \ 5y \ \frac{5^{y-1} -1}{y-1} $$ $$=5(1)\cdot \ \lim_{y\to 1} \frac{(5^{y-1} -1)’}{(y-1)’}$$ $$= 5\cdot\lim_{y\to 1} \frac{5^{y-1} \cdot \ln 5}{1}$$ $$=5\ln 5$$