Question #60ff3

Sep 3, 2017

Break it down as follows

Explanation:

${\sin}^{2} \frac{3 x}{{x}^{2} \cos x} = {\sin}^{2} \frac{3 x}{{x}^{2}} \cdot \frac{1}{\cos} x$
$= {\left(\sin \frac{3 x}{x}\right)}^{2} \cdot \frac{1}{\cos} x$
$= 9 \cdot {\left(\sin \frac{3 x}{3 x}\right)}^{2} \cdot \frac{1}{\cos} x$

Now, as $x \rightarrow 0$, $3 x \rightarrow 0$, and therefore
$\sin \frac{3 x}{3 x} \rightarrow 1$
This means that its square also goes to 1.

Since cosine is continuous at 0, $x \rightarrow 0$, $\cos x \rightarrow 1$

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