I am asked find the following limit
$$\lim_{\theta \rightarrow 0}\frac {\sin^2\theta}{\theta}$$
I recognize that $$\lim_{\theta \rightarrow 0}\frac{\sin\theta}{\theta}=1$$
But because I have $sin^2\theta$ in the numerator, I am left with…
$$\lim_{\theta \rightarrow 0}1(\sin\theta)$$
When I think about what this implies, I reason that the ratio of the opposite side over hypotenuse of the angle $\theta$ must approach approach zero, but for this to happen the opposite side would have a value of zero, which means the triangle formed would have no x component.
$$\lim_{\theta \rightarrow 0}1(\sin\theta)=0$$
Is my reasoning correct? Am I thinking about this question in a constructive manner?
Best Answer
$$ \lim_{\theta\to0} \frac{(\sin\theta)^2}{\theta} = \lim_{\theta\to0} \left(\sin\theta\frac{\sin\theta}{\theta}\right). $$ This is equal to $$ \left(\lim_{\theta\to0} \sin\theta\right)\left(\lim_{\theta\to0} \frac{\sin\theta}{\theta}\right) $$ if both of these last two limits exist.
Later note: "Exist" in this case means both are real numbers, not things like $\infty$ or $-\infty$.