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You're computing an improper integral (with limits on it) to see if something converges, and you're at the part where you have something like: $$\lim_{b\to\infty}\left(\frac{\tan^{-1}(b)^2}{2} - \frac{\tan^{-1}(1)^2}{2}\right).$$ Wouldn't it diverge because you have infinity in the equation?

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up vote 4 down vote accepted

Basically, i understand your question as "What is the limit $\lim_{b \to \infty} \arctan(b)$?" ($\arctan$ is just another notation for $\tan^{-1}$)

Well, $$\lim_{b \to \infty} \arctan(b) = \frac{\pi}{2}$$ How do know this? What does this mean?

To get a start, look at this plot of $\arctan(x)$. Doesn't it look like it tends to $\frac{\pi}{2}$ as $x$ gets larger? The mathematical reason for the limit being what it is, is that we can show, as $x$ gets larger and larger, $\arctan(x)$ gets closer and closer to $\frac{\pi}{2}$.

Another way to view this, in the case of $\arctan$ is considering what $\arctan$ actually is: the inverse of $\tan$ in the open intervall $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. What does $\tan(x)$ tend to as $x$ gets closer to $\frac{\pi}{2}$? That's right, infinity. (What is $\lim_{x \to -\infty} \arctan(x)$?)

As a simpler example of a function that does not tend to infinity as it's argument gets larger and larger, you can consider a constant function $f(x) = b$. What is $\lim_{x \to \infty} f(x)$? It is, of course, $b$, since $f(x)$ doesn't actually depend on $x$.

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so if it was arcsin then it would be 1 or would it be pi? – Cascadia Mar 17 '11 at 2:38
@Cascadia: $\arcsin x$ is only defined on the range $[-1,1]$, as those are the only values $\sin x$ can take. So you can't take $$\lim_{x \to \infty} \arcsin x.$$ However, $$\lim_{x \to 1} \arcsin x=\frac{\pi}{2}$$ – Ross Millikan Mar 17 '11 at 2:45

Not in this case, $\lim_{b\to\infty}\tan^{-1}(b)=\frac{\pi}{2}$. To see this, it helps to think of $\tan(x)$ as the slope of a ray from the origin rotating $x$ radians counterclockwise around the unit circle. So after travelling $\frac{\pi}{2}$ radians, the line from the origin is pointing straight up in the positive $y$ direction. The slope of this line is undefined, or $\infty$. So it makes sense that the limit above evaluates to what it does.

So in this particular case, if I'm interpreting your powers correctly, your expression evaluates to

$$\frac{(\pi/2)^2}{2}-\frac{\tan^{-1}(1)^2}{2}=\frac{\pi^2}{8}-\frac{\pi^2}{32}=\frac{3\pi^2}{32}$$ since $\tan^{-1}(1)=\frac{\pi}{4}$.

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You should show us the sequence you are working with. $tan^{-1}(\infty)$ is not well-defined because $\infty$ is not a real number; however, the limit of $tan^{-1}(x)$ as $x$ goes to $\infty$ is $\pi/2$.

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