Question

How do you compute the limit of `sin(7x)/sin(2x)` as `x->0`?

Answer 1

`lim_(x->0)frac sin(7x) sin(2x) =7/2`

Explanation:

Considering that:

`lim_(x->0) frac sin(alphax) (alphax) =1`

You can express:

`frac sin(7x) sin(2x) = 7x frac sin(7x) (7x) frac (2x) sin(2x) 1/(2x)`

`frac sin(7x) sin(2x) = 7/2 frac frac sin(7x) (7x) frac sin(2x) (2x)`

and then:

`lim_(x->0)frac sin(7x) sin(2x) = lim_(x->0)7/2 frac frac sin(7x) (7x) frac sin(2x) (2x) =7/2 *1/1 =7/2`