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infinite limits and e^x

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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.


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.

OK...  thanks!   That's what I suspected.  I think my solutions guide is wrong...

What does your solutions guide say? e^(-x) would approach infinity as x approaches negative infinity.