How do you find the limit of #(2-e^x)/(2+3e^x)# as x approaches infinity?

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Answer 1 · 2016-11-01T23:53:16

# lim_(x->oo)(2-e^x)/(2+3e^x) = -1/3#

If we look at the graph of #y=(2-e^x)/(2+3e^x)# we can see that it is clear that the limit exists, and is approximately #-1/3#

graph{(2-e^x)/(2+3e^x) [-10, 10, -2, 2]}

Now, As #x->oo# then #e^x->oo# ,but #e^-x->0#

So, it would be better if we could replace #e^x# with #e^-x#

# lim_(x->oo)(2-e^x)/(2+3e^x) = lim_(x->oo)((2-e^x))/((2+3e^x)) * e^-x/e^-x #

# :. lim_(x->oo)(2-e^x)/(2+3e^x) = lim_(x->oo)(e^-x(2-e^x))/(e^-x(2+3e^x)) #

# :. lim_(x->oo)(2-e^x)/(2+3e^x) = lim_(x->oo) (2e^-x-e^-xe^x) / (2e^-x+3e^-xe^x)#

# :. lim_(x->oo)(2-e^x)/(2+3e^x) = lim_(x->oo) (2e^-x-1) / (2e^-x+3)#

And, using #e^-x->0# as #x->oo# we have;

# lim_(x->oo)(2-e^x)/(2+3e^x) = (0-1) / (0+3) = -1/3#

Which is completely consistent with the above graph.