Hello! Limit help. Would be posted below. Thank you!

Question

mortonsalt · Answer

$$\lim_{x \rightarrow 0}\frac{1}{\arctan\frac{1}{2^x-1}}$$

anonymous · Answer

may you help me

anonymous · Answer

Let's try something like this: set $y=\arctan \dfrac{1}{2^x-1}$ so that $\tan y=\dfrac{1}{2^x-1}$, or $\log_2(\cot y+1)=x$.

The point of this is to rewrite the limit as
$$\lim_{x\to0}\frac{1}{\arctan\dfrac{1}{2^x-1}}=\lim_{y\to\color{red}{?}}\frac{1}{y}$$
Now $x\to0$ is the same as $\log_2(\cot y+1)\to0$, which must mean that $\cot y+1\to1$, which in turn suggests $y\to\dfrac{\pi}{2}$, but it could also mean $y\to-\dfrac{\pi}{2}$.

So for one approach value for $x$, namely $x\to0$, you have two different approach values for $y$, both of $\pm\dfrac{\pi}{2}$. What do you think that suggests about the limit?