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!
Best Answer
$\lim\limits_{x\to0}\dfrac{5^{1+\tan^2(x)}-5}{1-\cos^2(x)}=$
$=5\lim\limits_{x\to0}\dfrac{5^{\tan^2(x)}-1}{\sin^2(x)}=$
$=5\lim\limits_{x\to0}\left[\dfrac{5^{\tan^2(x)}-1}{\tan^2(x)}\cdot\dfrac1{\cos^2(x)}\right]=$
$=5\lim\limits_{x\to0}\dfrac{5^{\tan^2(x)}-1}{\tan^2(x)}\cdot\lim\limits_{x\to0}\dfrac1{\cos^2(x)}\underset{\overbrace{\text{by letting }y=\tan^2(x)}}{=}$
$=5\lim\limits_{y\to0}\dfrac{5^y-1}y\cdot\lim\limits_{x\to0}\dfrac1{\cos^2(x)}=$
$=5\cdot\ln5\cdot1=$
$=5\ln5\;.$
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 $x\to0\;.$