4
$\begingroup$

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?

  • 1
    The point is that the sine function is not defined as a ratio in calculus. It is defined either in terms of a power series, or in terms of a parametrization of the unit circle. If you are trying to figure out what $\sin(0)$ is equal to, then this is a basic value you should know and not something you figure out based on drawing pictures of triangles or circles. Either as a power series or as a parametrization of the unit circle, the fact that $\sin(0)=0$ should be evident and clear.2012-07-01

1 Answers 1

5

$ \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 $-\infty$.

  • 0
    @Michael Hardy, that clears things up. I can see now that limit will end up being 0.2012-07-01