# Continuity of arctan

looserlama

## Homework Statement

Let zn = Arg(-1 + i/n). Find limn→∞ zn

## Homework Equations

Definition of convergence of a sequence.

## The Attempt at a Solution

Well zn = Arg(-1 + i/n) = arctan(-1/n).

So it seems clear that limn→∞arctan(-1/n) = arctan(limn→∞ -1/n) = arctan(0) = $\pi$.

Which is true if arctan is continuous.

There's the problem, I don't know how to prove that arctan is continuous. It seems way more difficult that I expected. I can't seem to get anywhere with it.

How do you show it?

Or can anyone think of another way to find the limit?

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## Homework Statement

Let zn = Arg(-1 + i/n). Find limn→∞ zn

## Homework Equations

Definition of convergence of a sequence.

## The Attempt at a Solution

Well zn = Arg(-1 + i/n) = arctan(-1/n).

So it seems clear that limn→∞arctan(-1/n) = arctan(limn→∞ -1/n) = arctan(0) = $\pi$.

Which is true if arctan is continuous.

There's the problem, I don't know how to prove that arctan is continuous. It seems way more difficult that I expected. I can't seem to get anywhere with it.

How do you show it?

Or can anyone think of another way to find the limit?
Well, the derivative of arctan(x) is $\displaystyle \frac{1}{1+x^2}\,,$ which is continuous for all x.

If arctan(x) is differentiable, what does that say about its continuity?

looserlama
Yea if it is differentiable then it's continuous. But I'm guessing I can't just state that the derivative of arctan is 1/(1 + x2).

So I need to prove it.

I've seen the derivation of it, using the triangle, but I don't think that's a legitimate proof is it?

Wouldn't I have to use the definition of a derivative? And that seems really difficult.