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Question #17145

1 Answer

Oct 22, 2017

#lim_(x->0) x^n/e^x=0, n in NN#

Explanation:

We need to find #lim_(x->0) x^n/e^x, n in NN#. For this, we can use the following limit laws

#lim_(x->c) (f(x))/(g(x))=(lim_(x->c)f(x))/(lim_(x->c)(g(x))#

#lim_(x->c)[f(x)]^(p)=(lim_(x->c)f(x))^p#

#thereforelim_(x->0) x^n/e^x=(lim_(x->0) x^n)/(lim_(x->0)e^x)=(lim_(x->0) x)^n/(lim_(x->0)e^x)=0^n/1=0#

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