Question

Question #8200b

Answer 1

`lim_(x->-oo) xln(1+1/x) = 1`

Explanation:

Substitute `y=1/x`. When `x->-oo` then `y->0`, so:

`lim_(x->-oo) xln(1+1/x) = lim_(y->0) ln(1+y)/y = 1`

graph{xln(1+1/x) [-10, 10, -5, 5]}