# Limit $\lim_{x\to\infty} x(\arctan(a^2x)-\arctan(ax))$

I have a limit $$\lim_{x\to\infty} x(\arctan(a^2x)-\arctan(ax))$$ and I know the solution $$\frac{a-1}{a^2}$$, but I dont have any Idea, how to calculate this limit or at least how to start. Any idea?

• Do you know what is $\lim_{x \to \infty} \arctan(x)$? – devianceee Jun 2 at 21:24
• Are you allowed to use Taylor expansions? – user170231 Jun 2 at 21:25
• As long as I know $\pi/2$. Is there possibility, that it was an error in my solutions, which I get (only resoult without caluclation). @user170231, how do u Solve with Taylor? – Vid Jun 2 at 21:29

I assume that we are near $$+\infty$$.

If $$a=0$$, the limit is zero.

If $$a<0$$ the limit is$$+\infty(\frac{\pi}{2}-(-\frac{\pi}{2}))=+\infty$$

If $$a>0$$, then we use the well-known identity, for $$X>0 \;:$$

$$\arctan(X)=\frac{\pi}{2}-\arctan(\frac 1X)$$

So, we want $$\lim_{x\to+\infty}x\Bigl(\arctan(\frac{1}{ax})-\arctan(\frac{1}{a^2x})\Bigr)$$

we use the fact that, near $$+\infty$$ $$\arctan(\frac 1X)=\frac 1X(1+\epsilon(X))$$

thus $$\arctan(\frac{1}{ax})-\arctan(\frac{1}{a^2x})=$$ $$\frac 1x(\frac 1a-\frac{1}{a^2})+\frac 1x\epsilon(x)$$

the limit is then $$\frac 1a-\frac{1}{a^2}=\frac{a-1}{a^2}$$