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.