Limit of e^-x as x approaches infinity vs approaching negative infinity?

Question

Why does the lim of e^-x as x approaches +infinity equal 0 but the lim of e^-x as x approaches - infinity equal infinity??

Shouldn't they both just be 1/e^x? Wouldn't this mean they should both equal 0?

Answer 1

Think of the sequence e^-1, e^-10, e^-100, e^-1000, etc. The terms get smaller and smaller.



Now think of the sequence e^1, e^10, e^100, e^1000, etc. The terms get bigger and bigger.

Answer 2

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RE:

Limit of e^-x as x approaches infinity vs approaching negative infinity?

Why does the lim of e^-x as x approaches +infinity equal 0 but the lim of e^-x as x approaches - infinity equal infinity??



Shouldn't they both just be 1/e^x? Wouldn't this mean they should both equal 0?

Answer 3

First off, I think you would already know but I'm saying it anyway. Limit essentially means the y-value of the function at the point that x approaches a certain value, like 5. If you plug in a very large number, like 1 million, for x, you get a number that is very small (1/e^1mil), hence the limit approaches 0. Now, plug in a very large NEGATIVE value for x ( -1 million) and you will see that it becomes e^(-(-1mil.) = e^1mil, an extremely large number, hence approaching positive infinity. If you graph the function using a graphing calculator or some online grapher, you'll see how it works. Hope this helps...

Answer 4

The number e is 2.718, which is very close to 3. So, to visualize the graph of e^x, we can assume it is close to the graph of 3^x. When looking at the graph of 3^x, and you take the limit as x -> negative infinity, you can see that the graph asymptotes at x=0, which means the limit of 3^x as x-> negative infinity is 0. This implies that the limit of e^x as x-> negative infinity is ALSO 0. In which case, you can plug in 0 for e^x and end up with 1/3 as your final answer :)

Answer 5

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e^x go to 0 if x-> - infinity then limit 2/(0+6 )= 1/3

Answer 6

I would like to direct you to a graph of e^x.



http://content.tutorvista.com/maths/content/us/class11maths/chapter17/images/img40.gif



Or just google "e to the x"



As lim e^-x approaches infinity, you have e^-infinity which brings you all the way to the left of the graph, which you can see is approaching zero.



As lim e^-x approaches negative infinity you have e^infinity which is going all the way to the right of the graph which as you can see is approaching infinity.

Answer 7

e^-infinity = 1/e^infinity or 1/infinity which = 0.

e^--infinity = e^infinity = infinity

Answer 8

E To The X