# How to show \$g(x)=frac{4^x+x^2-ln(2)cdot x-1}{tan(2x)}\$ is continuous at \$x = 0\$ for \$g(0) := ln(sqrt{2})\$

\$g(x)\$ is defined on the Intervalls \$(frac{-pi}{2}, 0) cup(0,frac{pi}{2})\$

I’ve tried doing it by using L’Hôpital but \$sin(x)\$ gets into the denominator and I can’t get rid off it.

Hint: By writing \$tan(cdot)=frac{sin(cdot)}{cos(cdot)}\$ and rearranging terms, we can show that:

\$\$frac{4^x+x^2-ln(2) x-1}{tan(2x)}=cos(2x)left(color{blue}{frac{4^x-1}{sin(2x)}}right)+cos (2x)left(x-ln 2right)left(color{red}{frac{x}{sin(2x)}}right)\$\$

So now all we need to find is \$limfrac{4^x-1}{sin(2x)}\$ and \$limfrac{x}{sin(2x)}\$.

Hint 2:

We can simplify the problem further by stating \$frac{x}{sin(2x)}=frac{1}{2}cdotfrac{2x}{sin(2x)}\$ and using the substitution \$2xmapsto u\$. so the limit becomes:

\$\$lim_{xto0}frac{x}{sin(2x)}=frac{1}{2}:color{green}{lim_{uto0}frac{u}{sin u}}\$\$

This limit is very famous and can be proven geometrically. We can then state \$limfrac{4^x-1}{sin(2x)}=frac{2^{2x}-1}{sin(2x)}=left(frac{2^{2x}-1}{2x}right)left(frac{2x}{sin(2x)}right)\$. We can simplify the bracketed term on the left hand side as \$frac{2^{2x}-1}{2x}=frac{e^{(2ln2) x}-1}{2x}\$. With the substitutions \$2 x mapsto u\$ and \$(2ln2) x mapsto v\$, we can split the limit in two to show:
\$\$
lim_{xto0}frac{4^x-1}{sin(2x)}=ln2:color{brown}{lim_{vto 0}frac{e^v-1}{v}}:color{green}{lim_{uto 0}frac{u}{sin u}}\$\$

There are many ways to prove both of these limits, including ways that avoid l’Hôpital’s rule.