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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?

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  • $\begingroup$ Do you know what is $\lim_{x \to \infty} \arctan(x)$? $\endgroup$ – devianceee Jun 2 at 21:24
  • $\begingroup$ Are you allowed to use Taylor expansions? $\endgroup$ – user170231 Jun 2 at 21:25
  • $\begingroup$ 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? $\endgroup$ – Vid Jun 2 at 21:29
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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 $$\frac 1a-\frac{1}{a^2}=\frac{a-1}{a^2}$$

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