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.

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