Inverse trignometric functions and Mean value theorem

Question

I know my answer is wrong but I can't find my mistake or I don't have a profound understanding of the 2nd form of MVT. Can anyone guide me please

Answer 1

The MVT theorem states that there exists $\theta\in (0,1)$ such that

$$\arctan(x+h)-\arctan x=\dfrac{h}{1+\theta^2h^2}.$$

Now, if $x=0$ then we have

$$\arctan h-\arctan 0=\dfrac{h}{1+\theta^2h^2}.$$ Writing $h=x$ we have

$$\arctan x-\arctan 0=\dfrac{x}{1+\theta^2x^2}.$$ Since $\arctan 0=0$ it is

$$\arctan x=\dfrac{x}{1+\theta^2x^2}.$$ So

$$1+\theta^2x^2=\dfrac{x}{\arctan x}.$$ Thus

$$\theta^2=\dfrac{1}{x\arctan x}-\dfrac{1}{x^2}$$ or, equivalently

$$\theta=\sqrt{\dfrac{1}{x\arctan x}-\dfrac{1}{x^2}}.$$

Finally, you need to compute $$\lim_{x\to 0}\left(\dfrac{1}{x\arctan x}-\dfrac{1}{x^2}\right)=\lim_{x\to 0}\dfrac{x-\arctan x}{x^2\arctan x}.$$