Finding asymptotes of g(x) = x/e^(x-1): vertical is none, but horizontal...?

[adeenah]

Given: g(x)=x/e^(x-1)

Vertical asymptotes: none

Horizontal: Not sure?
I tried to find the limit as x-->+/- infinity but am confused

 

[stapel]

Given: g(x)=x/e^(x-1)

Vertical asymptotes: none

Horizontal: Not sure?
I tried to find the limit as x-->+/- infinity but am confused

What did you do when you "tried"? At what point did you get confused? You know that x increases linearly, while e^(x-1) increases exponentially. Which is "stronger"? And you took the derivative and considered the First Derivative Test. What did you conclude from this?

Please be complete. Thank you!

 

[adeenah]

What did you do when you "tried"? At what point did you get confused? You know that x increases linearly, while e^(x-1) increases exponentially. Which is "stronger"? And you took the derivative and considered the First Derivative Test. What did you conclude from this?

Please be complete. Thank you!

I considered: as x->infinity, e^(x-1) ->infinity & as x->-infinity, e^(x-1) ->0
so lim as x->infinity of g(x)=0 because e^(x-1) is "stronger" than x?
meaning a large number divided by an even larger number approaches 0
so y=0 would be a horizontal asymptote

and lim as x->-infinity g(x)=-infinity? because e^(x-1) is approaching a small positive number, while the numerator approaches negative infinity?

The first derivative test showed that there is a local minimum at g(1)=1, how does this connect with finding asymptotes?

Thank you!

 

[stapel]

I considered: as x->infinity, e^(x-1) ->infinity & as x->-infinity, e^(x-1) ->0
so lim as x->infinity of g(x)=0 because e^(x-1) is "stronger" than x?
meaning a large number divided by an even larger number approaches 0
so y=0 would be a horizontal asymptote

I agree.

and lim as x->-infinity g(x)=-infinity? because e^(x-1) is approaching a small positive number, while the numerator approaches negative infinity?

I agree with your limit value.

The first derivative test showed that there is a local minimum at g(1)=1...

Um... You should have found a critical point at x = 1, but I don't think it's a minimum...