Question #60ff3

1 Answer
Sep 3, 2017

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.