# Proving a limit

1. Sep 30, 2006

### bluevires

Hey guys

I'm having trouble proving this limit using delta-epsilon definition
I would appreciate some help if possible

$$\lim_{x\rightarrow \infty} \arctan(x)=\frac{\pi}{2}$$

I know that in order for the statement to be true,
Assuming
Epislon>0
Then |f(x)-L|<Epsilon for x> N

but i havn't had much experience working with trignometric functions, so I don't know how should I set my N equals to, and how could I convert that to |f(x)-L|< Epsilon

Last edited: Sep 30, 2006
2. Oct 1, 2006

### Mindscrape

Do you have to use a delta-epsilon definition to solve your problem? Just thinking about the graph of arctan(x) the answer is obvious what happens as x approaches infinity. It seems like a stupid question if you can just come up with the answer in a couple second.

3. Oct 1, 2006

### Hurkyl

Staff Emeritus
The point of a calculus class is to learn calculus. If you can't use calculus to compute something "obvious", then how are you going to use calculus to compute something that isn't "obvious"? :tongue:

bluevires -- why not simply solve the inequality

$$| \arctan x - \pi / 2 | < \epsilon$$

for x, to figure out what you should use for N?

4. Oct 1, 2006

### Mindscrape

Yeah, but the point of any math class should be to allow for any viable method to be used, and for math to be open.

5. Oct 1, 2006

### Hurkyl

Staff Emeritus
No, the point of a math class is to learn math.

Learning to use alternate methods is certainly a part of math, and is a good thing, but using alternate methods at the expense of learning the subject you're supposed to be learning is bad.

(I often advise doing the same problem multiple ways, if you can!)

And, a big part of mathematics is being able to back up your intuition with rigor when appropriate -- it was intuition that told you what the value of the limit should be, but what if someone didn't have as much faith in your intuition? (e.g. a co-worker... a teacher... your boss...)

6. Oct 1, 2006

### bluevires

thank you hurkyl, problem solved.