Question

How do you evaluate the limit `(sin(2x))/(2x^2+x)` as x approaches `0`?

Answer 1

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

Explanation:

Use the fact that:

`lim_(t->0) sin(t)/t = 1`

Hence:

`lim_(x->0) sin(2x)/(2x^2+x) = lim_(x->0) (sin(2x) -: (2x))/((2x^2+x) -: (2x))`

`color(white)(lim_(x->0) sin(2x)/(2x^2+x)) = lim_(x->0) (sin(2x) -: (2x))/(x+1/2)`

`color(white)(lim_(x->0) sin(2x)/(2x^2+x)) = 1/(1/2)`

`color(white)(lim_(x->0) sin(2x)/(2x^2+x)) = 2`

Answer 2

`2`

Explanation:

`sin(2x)/(2x^2+x)=2(sin(2x)/((2x)))1/(2x+1)`

so

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