Question

How do you evaluate the limit Sin^2(3x)/x^3-x as x approaches 0?

Answer 1

`0`

Explanation:

Your limit: `\lim _ ( x \rightarrow 0 ) (sen^2(3x) )/( x ^ ( 3 ) - x )`

Using de l'Hôpital rule will be: `\lim _ ( x \rightarrow 0 ) ( 6 sin ( 3 x ) cos ( 3 x ) )/( 3 x ^ ( 2 ) - 1 )`

For `x -> 0` will be:

• `sin(3x) -> 0`
• `cos(3x) -> 0`
• `3x^2 -> 0`

So: `lim_(x->0)0/-1=0`