Prove that $\lim_{x\rightarrow \infty}\tan^{-1}x=\pi/2$

Question

I have recently started studying limits when I came across this question:

Prove that $\lim_{x\rightarrow \infty} \tan^{-1}x=\dfrac{\pi}{2}$ using $\epsilon-\delta$ approach.

This question was given as exercise and I have approached it in this way:

My Appraoch:

Suppose $\Big|\tan^{-1}x-\dfrac{\pi}{2}\Big|<\epsilon$

Now $x>0$ since $x\rightarrow \infty$, this means that the maximum value of $\Big|\tan^{-1}x-\dfrac{\pi}{2}\Big|$ is $\dfrac{\pi}{2}$.

When $\epsilon>\dfrac{\pi}{2}$, then the $\epsilon$-inequality is valid for any $x>0$

When $\epsilon<\dfrac{\pi}{2}$, then: $$\tan^{-1}x>\dfrac{\pi}{2}-\epsilon$$ $$x>\tan\Big(\dfrac{\pi}{2}-\epsilon\Big)$$

Thus for any $0<\delta\leq\tan\Big(\dfrac{\pi}{2}-\epsilon\Big)$, we would have $x>\delta$ which implies that $\Big|\tan^{-1}x-\dfrac{\pi}{2}\Big|<\epsilon$.

$\therefore \lim_{x\rightarrow \infty} \tan^{-1}x=\dfrac{\pi}{2}$

Please help me verify this solution and please offer some suggestions.

THANKS

Answer 1

If $x>0$, then

$$\arctan(x)=\pi/2-\arctan(1/x)\tag1$$

In This Answer, I showed using elementary analysis only that for $x>0$

$$\arctan(x)\le x\tag2$$

Using $(1)$ and $(2)$ reveals that for any given $\varepsilon>0$

$$\begin{align}\left|\arctan(x)-\frac \pi2\right|&=\left|\arctan\left(\frac1x\right)\right|\\\\ &\le \frac1x\\\\ &<\varepsilon \end{align}$$

whenever $x>\frac1\varepsilon$. And we are done!