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

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Answer 1 · 2016-11-03T01:01:44

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

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

graph{(e^x-x^2)/(e^x+x) [-5, 15, -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)(e^x-x^2)/(e^x+x) = lim_(x->oo)((e^x-x^2)/(e^x+x)) * e^-x/e^-x #

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

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

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

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

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

Which is completely consistent with the above graph.

Answer 2 · 2016-11-03T01:29:32

1

Use iteratively L'Hospital's rule, until the indeterminate form #oo/oo#

disappears..

#lim x to oo ((e^x-x^2)/(e^x+x))#

#=lim x to oo (((e^x-x^2)')/((e^x+1)'))#

#=lim x to oo (((e^x-2)')/((e^x)'))#

#lim x to oo (1-2e^(-x))#

#=(1-0)=1#