Prove that arctanx has to be less than x

[AdityaDev]

Homework Statement

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

Homework Equations

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?

 

[PeroK]

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

 

[AdityaDev]

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]

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]

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 Statement

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

Homework Equations

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:
$$\arctan(x) \begin{cases} < x,& x > 0\\ > x, & x < 0<br /> \end{cases}$$
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 -.

 

[AdityaDev]

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:
$$\arctan(x) \begin{cases} < x,& x > 0\\ > x, & x < 0<br /> \end{cases}$$
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+

 

[AdityaDev]

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

 

[SammyS]

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 !)

 

[Ray Vickson]

Look at Post #5 from vela .

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

Actually, exactly equal to one.

 

[AdityaDev]

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]

No. its x->0+. Using graph, the arctanx curve starts coming below y=x.

Do you understand limits ?

 

[AdityaDev]

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]

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.