Question

Question #60ff3

Answer 1

Break it down as follows

Explanation:

`sin^2(3x)/(x^2cosx) = sin^2(3x)/(x^2)*1/cosx`
`= (sin(3x)/x)^2*1/cosx`
`= 9*(sin(3x)/(3x))^2*1/cosx`

Now, as `x rarr 0`, `3x rarr 0`, and therefore
`sin(3x)/(3x) rarr 1`
This means that its square also goes to 1.

Since cosine is continuous at 0, `x rarr 0`, `cosx rarr 1`

Put this together, and you should arrive at an overall limit of 9.