Limit of Trig without using L'Hopitals rule

Question

I am confuse with the part of sin^2. I am not sure if there is a trig identity to simplify. I was thinking of trying to rearranging the equation so that it would be similar to sin(x)/x to solve it. Which direction do I go since I have come to road blocks with both?

Answer 1

$$\begin{array}{rcl} \displaystyle \lim_{x\to0} \frac{\sin^2(3x)}{x^2\cos x} &=& \displaystyle \lim_{x\to0} \frac{\sin(3x)}{3x} \cdot \frac{\sin(3x)}{3x} \cdot \frac{9}{\cos x} \\ &=& 1 \cdot 1 \cdot 9 \\ &=& 9 \end{array}$$

Answer 2

Drop the cosine and use a change of variable to absorb the $3$ under the sine:

$$\lim_{x\to0}\frac{\sin^2(3x)}{x^2\cos x}=\lim_{u\to0}\frac{\sin^2(u)}{\left(\dfrac u3\right)^2}=9\left(\lim_{u\to0}\frac{\sin u}u\right)^2.$$