Question

DIVERGE OR CONVERGE? CONVERGE FOR SURE, but....
Question:
http://i1084.photobucket.com/albums/j409/QRAWarrior/MATA36/MATA36-Question4Q.png

Attempt: (must see both parts to make sense)
Part a:
http://i1084.photobucket.com/albums/j409/QRAWarrior/MATA36/MATA36-Question4A.png

Part b:
http://i1084.photobucket.com/albums/j409/QRAWarrior/MATA36/MATA36-Question4B.png

Answer 1

I am having trouble evaluating the limit. I know for sure this is convergent.

Answer 2

@SmoothMath
@eliassaab
@lalaly
@jim_thompson5910
@asnaseer

Answer 3

@AccessDenied

Answer 4

@cwrw238

Answer 5

@QRAwarrior

Answer 6

@asnaseer
@mahmit2012
@amistre64
@KingGeorge
@jim_thompson5910
@cwrw238
@dpaInc
@eliassaab
@SmoothMath

Answer 7

Anybody have any idea where I may have messed up?

Answer 8

Sorry I should have made that clear

Answer 9

@JamesJ

Answer 10

@Chlorophyll
@phi
@Mertsj

Answer 11

Wolfram can do this integral
http://www.wolframalpha.com/input/?i=integral+16+exp%281%2Fx%29+x%5E%28-3%29+dx+

click on "show steps" to get the details

Answer 12

I am pretty sure I got past the part of evaluating the integral. Had I not, I would not be at evaluating the limit!

Answer 13

@phi

Answer 14

@JamesJ
@amistre64
@FoolForMath
@mahmit2012
@SmoothMath
@Mertsj
@jim_thompson5910
@asnaseer
@AccessDenied

Answer 15

We can turn to wolfram for more insight
http://www.wolframalpha.com/input/?i=limit+x-%3E-1+%2816+exp%281%2Fx%29%28x-1%29%2Fx%29

Answer 16

Wow. I forgot I could use wolfram to evaluate the limit!!

Answer 17

QRA, REALLY? You're going to mention me in this question THREE TIMES? So obnoxious.

Answer 18

@phi You have been mistaken with what you put as the limit approaches. It is 0 to the left, not 1 to the left.

Answer 19

@amistre64 Hey amistre, would you mind helping me with this convergence problem?

Answer 20

Alright this is the PROPER limit output for Wolfram, but the problem is that it does not show the steps!
http://www.wolframalpha.com/input/?i=limit+x-%3E0-+%28e%5E%281%2Fx%29%29%2Fx

Answer 21

When x is <0 that puts exp(1/x) in the denominator. It goes to inf "faster" than x goes to 0