What is the horizontal asymptote of y=#3e^(-1/x)# ?

Question

I plugged in positive and negative infinity into x and got #3e^0# which gave me a horizontal asymptote of 3, is this correct?

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Answer 1 · 2018-01-23T21:32:15

#y=3#

I agree that #y=3# is the horizontal asymptote.

Since #-1/x# approaches 0 as #x# tends toward both positive and negative infinity, #3e^(-1/x)# tends toward #3e^0=3#.