How do you find the horizontal asymptote for #g(x)=e^x-2#?

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Answer 1 · 2016-01-14T01:24:06

Evaluate x at both #+-oo#

#""_(xrarroo)^"lim"(e^x-2)=oo#

So, the right limit, as #xrarroo#, has no horizontal asymptote.

However, the left limit does have an asymptote:

#""_(xrarr-oo)^"lim"(e^x-2)=0-2=-2#

So, the left limit, as #xrarr-oo#, has horizontal asymptote #=-2#.

hope that helped

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