Evaluating the Limit of e^tan x - How to Solve?

[phil ess]

Homework Statement

Evaluate the following limit. lim as x approaches Pi/2 ⁺ of e^{tan x}.

Homework Equations

None.

The Attempt at a Solution

Well I've graphed this and I know it approaches 0, but I don't know how to actually solve this. And I'm fairly sure I'm not allowed to just substitute a number close to Pi/2, unless that's the only way. Thanks!

 

[=Lawrence=]

The limit is zero. Actually, you can go to the table on your calculator (2nd, graph) and type in numbers that are extremely close and that will work. The limit is what it approaches, in this case from the right, and it's 0.

 

[Dick]

The limit of tan(x) as x decreases to pi/2 is -infinity. To show that tan(x)=sin(x)/cos(x) and you probably know their limit around pi/2. Or just look at a graph of tan(x). Then just look at a graph of e^x as x->-infinity.

 

[=Lawrence=]

The limit of tan(x) as x decreases to pi/2 is -infinity. To show that tan(x)=sin(x)/cos(x) and you probably know their limit around pi/2. Or just look at a graph of tan(x). Then just look at a graph of e^x as x->-infinity.

Are you disagreeing with me? I can't tell what point you're trying to get across.

The limit of e^tan(x) as x approaches pi/2 from the right is zero.

 

[Dick]

No, I'm agreeing with you. I'm just trying to say how to describe the problem without a calculator.

 

[phil ess]

Ok thanks. I already knew the answer, and I can explain it conceptually. I just thought there was a more, mathematical?, way of showing the answer.

 

[Dick]

You can do epsilons and deltas if you want. I don't think that's necessary.

 

[BoundByAxioms]

Here is an informal way to think about it. The limit as x approaches pi/2+ of tan(x)= -infinity. e^(-infinity) = 1/(e^infinity) = 0.

 

[HallsofIvy]

e^x and tan(x) are both continuous functions. tan(\pi/2)= 0 and e⁰= 1. Drum roll.

 

[Dick]

e^x and tan(x) are both continuous functions. tan(\pi/2)= 0 and e⁰= 1. Drum roll.

tan(pi/2)=0? tan(x) continuous?