Question

Question #17145

Answer 1

`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`