Question

How do you evaluate the limit `lim(x^2+xcosx)/(2x)dx` as `x->0`?

Answer 1

`lim_(x->0)(x^2+xcosx)/(2x) = 1/2`

Explanation:

Simplifying, we can see that the function is continuous for `x=0`, in fact:

`lim_(x->0)(x^2+xcosx)/(2x) = lim_(x->0)(cancel(x)(x+cosx))/(2cancel(x))`

`lim_(x->0)(x^2+xcosx)/(2x) = lim_(x->0) (x+cosx)/2 = (0+1)/2 = 1/2`

graph{(x^2+xcosx)/(2x) [-10, 10, -5, 5]}