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A while back I ran into a problem in which I had to analyze the graph of $f(x) = ( \arctan x )^2$.

I was fine until I had to evaluate the limit of the function as is approaches infinity to determine if there were asymptotes. I did not know how to solve this algebraically or analytically so I proceeded to try graphically. Still, I was having trouble so I decided to ask a Calculus teacher at my school, but even then the teacher proceded to tell me that "you don't need to do that" and "this is past what we are doing in class" Obviously I am not just going to stop self studying math, I have been doing it for about 2 months now and I am not planning on stopping just because I am struggling with one concept so far. So after a while of trying to find a way to solve this I decided to move on. To my amusement one of these problem appeared again and now I need to understand this if I am ever going to proceed with my studies. The problem is an example from my textbook. The question says "Apply the Integral Test to the series $1/(n^2+1)$. I integrate and get the limit as b approaches infinity of [arctanb-arctan1]. At that point I am stuck.... The answer is given to be $\pi/2-\pi/4$ which is equal $\pi/4$. Please if anyone can shed some light on how to come about these answers.

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Draw a right triangle showing an "adjacent" side and an "opposite" side. Let the length of the adjacent side by $1$. Then since $\tan = \dfrac{\mathrm{opp}}{\mathrm{adj}}$, the tangent of the angle is equal to the opposite side.

Therefore the arctangent of the opposite side is the angle.

What does the angle approach as the length of the "opposite" side goes to $\infty$?

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The trick to the limits of the inverse trigonometric functions is knowing where they come from.

Let me start with the example of the function $f(x)=cosx$, you know that the functions domain is all real numbers and the range is $-1\le y \le1$. To find the inverse of this function, we have to start by restricting the domain, in this case we want the piece of the function from $0\le x \le\frac{\pi}{2}$. Then, we reflect the function across $y=x$, making the range into the domain. Thus, the domain of the function $f(x)=arctanx$ is $-1\le x \le1$ and the range is $0\le y \le\frac{\pi}{2}$. So, the limit as this function approaches $1$ is $\frac{\pi}{2}$

This method can also be applied to the function $f(x)=tan(x)$, in that the domain is all real numbers excluding the asymptotes at values of $\frac{\pi}{2}$ and the range is all real numbers. By using the same method as above, we find that the domain of $f(x)=arctan(x)$ is all real numbers, and the range is $-\frac{\pi}{2}\le y \le\frac{\pi}{2}$

So, the $\lim_{x \to +\infty}arctan(x)=\frac{\pi}{2}$

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