# Question about using an integral to see if something converges

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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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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