Question

Question #969c8

Answer 1

See below.

Explanation:

e)
`(x^2-16)/(x^2+4)=(x^2+4-20)/(x^2+4)=1-20/(x^2+4)`

`lim_(x->oo)(x^2-16)/(x^2+4)=1-lim_(x->oo)20/(x^2+4)=1`

f)
`lim_(x->0)x^3/(2x+1)=(lim_(x->0)x^3)(lim_(x->0)1/(2x+1))=0 cdot 1=0`