# Prove that arctanx has to be less than x

## Homework Statement

Prove that :
$$\lim_{x\to 0}\frac{tan^{-1}x}{x}<1$$
And
$$\lim_{x\to 0}\frac{sin^{-1}x}{x}>1$$

None

## The Attempt at a Solution

I have to prove that arctanx has to be lesser than x.
It's derivative is 1 at x=0 and keeps decreasing as x increases. So it's slope decreases from 1.
Also, at x=0, arctanx=x. So the curve which initially touched y=x goes below it. So arctanx<x.
Is this correct?

## Answers and Replies

PeroK
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Why not think about tan(x) and sin(x)?

I know that sinx<x.
So x<arcsinx..... By taking arcsin on LHs and RHS.
Thank you.
Is my method in post 0 correct?

PeroK
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It might be neater to consider the functions

f(x) = x - arcsin(x)

g(x) = x - arctan(x)

Shouldn't the limits be ##\le## and ##\ge##?

Why not use L'hopital to calculate them?

vela
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## Homework Statement

Prove that :
$$\lim_{x\to 0}\frac{tan^{-1}x}{x}<1$$
And
$$\lim_{x\to 0}\frac{sin^{-1}x}{x}>1$$
You can't prove these because they're false. The limit is a single number, i.e.,
$$\lim_{x \to 0}\frac{\tan^{-1}x}{x} = 1.$$

Ray Vickson
Homework Helper
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## Homework Statement

Prove that :
$$\lim_{x\to 0}\frac{tan^{-1}x}{x}<1$$
And
$$\lim_{x\to 0}\frac{sin^{-1}x}{x}>1$$

None

## The Attempt at a Solution

I have to prove that arctanx has to be lesser than x.
It's derivative is 1 at x=0 and keeps decreasing as x increases. So it's slope decreases from 1.
Also, at x=0, arctanx=x. So the curve which initially touched y=x goes below it. So arctanx<x.
Is this correct?

The statements you are trying to show are both false. If you think ##\lim_{x \to 0} \arctan(x)/x < 1##, how much < 1 do you think it is? Is it < 0.9? Is it < 0.99?

Also: arctan(x) is not < x; in fact:
$$\arctan(x) \begin{cases} < x,& x > 0\\ > x, & x < 0 \end{cases}$$
Just plot the two functions to see that this is true.

You should likewise check ##\sin(x)## against ##x## for small ##x## of both signs, + and -.

The statements you are trying to show are both false. If you think ##\lim_{x \to 0} \arctan(x)/x < 1##, how much < 1 do you think it is? Is it < 0.9? Is it < 0.99?

Also: arctan(x) is not < x; in fact:
$$\arctan(x) \begin{cases} < x,& x > 0\\ > x, & x < 0 \end{cases}$$
Just plot the two functions to see that this is true.

You should likewise check ##\sin(x)## against ##x## for small ##x## of both signs, + and -.
sorry... its x->0+

so you can prove it for x->0+ by using LH rule directly and substituting 0+.

SammyS
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so you can prove it for x->0+ by using LH rule directly and substituting 0+.
Look at Post #5 from vela .

The limits you are referring to are all exactly equal to zero one .

(Thanks to RGV !)

Last edited:
Ray Vickson
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Look at Post #5 from vela .

The limits you are referring to are all exactly equal to zero.

Actually, exactly equal to one.

Look at Post #5 from vela .

The limits you are referring to are all exactly equal to zero one .

(Thanks to RGV !)
No. its x->0+. Using graph, the arctanx curve starts coming below y=x.

SammyS
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No. its x->0+. Using graph, the arctanx curve starts coming below y=x.
Do you understand limits ?

yes. its a homework question to prove that $$\lim_{x\to 0}\sqrt{\frac{tan^{-1}x}{x}-\frac{sin^{-1}x}{x}}$$ does not exist and its given in the solution that ##\frac{tan^{-1}x}{x}<1## while ##\frac{sin^{-1}x}{x}>1##, when x->0.
If it is wrong, even then ##\frac{tan^{-1}x}{x}<\frac{sin^{-1}x}{x}## when x->0 (actually 0+).

SammyS
Staff Emeritus
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yes. its a homework question to prove that $$\lim_{x\to 0}\sqrt{\frac{tan^{-1}x}{x}-\frac{sin^{-1}x}{x}}$$ does not exist and its given in the solution that ##\frac{tan^{-1}x}{x}<1## while ##\frac{sin^{-1}x}{x}>1##, when x->0.
If it is wrong, even then ##\frac{tan^{-1}x}{x}<\frac{sin^{-1}x}{x}## when x->0 (actually 0+).
That's quite different from the original post.

For 0 < x ≤ 1 , ## \displaystyle\ \frac{\tan^{-1}x}{x}-\frac{\sin^{-1}x}{x} <0##

Taking the square root gives an imaginary result.