Question

Question #389da

Answer 1

2

Explanation:

If the problem is `lim_(x to 0)sin(2x)/(2x^2+x)` then direct substitution gives `0/0` so we can use L'Hopital's Rule:

`lim_(x to 0)sin(2x)/(2x^2+x) = lim_(x to 0) (d/dxsin(2x))/(d/dx(2x^2+x))`

`=lim_(x to 0) (2cos(2x))/(4x+1) = 2/1=2`