WyzAnt.comI understand that as x approaches (+) infinity, e^x = infinity.
I also understand that as x approaches (-) infinity, e^x = 0. The graph makes sense.
So, as x approaches (-) infinity, -e^x = infinity... Is this true?
infinite limits and e^x
WyzAnt Tutors: Sign In to answer this question.
1 Answer
No--because you have not changed the argument of the function at all (the exponent on top of the e), you are going to get 0 again. It's as if you multiplied both sides of the second equation by -1, giving you -1 * 0 = 0
Another way to look at it is to recall that if you multiply a function by -1 it is reflected about the x-axis. If you look at e^x reflected in such a way you can see that as you go towards -infinity you again get 0.
Comments
You could add that, as x approaches -infinity, e^x approaches 0+ (zero from the positive side of the y-axis) and -e^x approaches 0- (zero from the negative side of the y-axis).
More simply, ie from a graph, as x approaches negative infinity, e^x approaches zero from above, and -e^x approaches zero from below.
- Daniel O. 11/19/2012OK... thanks! That's what I suspected. I think my solutions guide is wrong...
- John R. from Apex, NC 11/19/2012What does your solutions guide say? e^(-x) would approach infinity as x approaches negative infinity.
- Daniel O. 11/19/2012