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

Do you want to show for all $\epsilon$ there is $X$ such that if $x>X$ then $|\tan^{-1}x-\dfrac \pi 2|<\epsilon$ — J. W. Tanner, Aug 28 '20 at 03:28
If you're trying to show $|\arctan x-(\pi/2)|<\epsilon$, then you can't start a proof by assuming $|\arctan x-(\pi/2)|<\epsilon$. — Gerry Myerson, Aug 28 '20 at 03:38
Also, for limits at infinity, epsilon-delta approach doesn't make sense, as it's impossible for the distance from $x$ to infinity to be less than delta. You need an $X$-epsilon approach, as in @J.W.Tanner comment. — Gerry Myerson, Aug 28 '20 at 03:40
@GerryMyerson I'm sorry for I may have misstated the approach but that was exactly what I was trying to do. — Devansh Kamra, Aug 28 '20 at 03:41

score 2 · Accepted Answer · answered Aug 28 '20 at 03:47

2

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!

answered Aug 28 '20 at 03:47

Mark Viola

179,405

A great approach indeed. Thanks for help (+1) – Devansh Kamra Aug 28 '20 at 03:51
You're welcome. My pleasure. – Mark Viola Aug 28 '20 at 04:01

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

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