How do you find the limit of #(e^-x-e^(-x/2))/sqrt(e^x+1)# as #x->oo#?

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Answer 1 · 2017-04-09T19:05:00

The limit is #0#. The initial form is not indeterminate, but it may be helpful to rewrite the expression.

This limit has initial form #0/oo# which is not indeterminate.

Rewrite it as

#(e^-x-e^(-x/2))*1/sqrt(e^x+1)#.

As #xrarroo#, this expression goes to

#(0-0)*0 = 0#

Note

If #lim_(xrarroo)f(x) = 0# and #lim_(xrarroo)g(x) = oo#, then

since (in this case) #lim_(xrarroo)1/g(x) = 0# we will always have

#lim_(xrarroo)f(x)/(g(x)) = 0#