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
    $\begingroup$ The sine function in calculus is not defined in terms of ratios of opposite sides and hypothenuses; it's defined in a different manner. Rather, you should use the fact that the sine function is continuous, so you can compute the limit as $\theta\to 0$ of $\sin\theta$ simply by plugging in $\theta=0$. $\endgroup$ – Arturo Magidin Jul 1 '12 at 19:15
  • $\begingroup$ @ArturoMagidin, I drew a picture of the unit circle and determined that for the ratio of the sin function to equal 0, the opposite side would have to be 0, which seems to be correct. I am wondering why the trig functions cannot be defined this way in calc, and in what way they are correctly defined? $\endgroup$ – Kurt Jul 1 '12 at 20:18
  • 2
    $\begingroup$ 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. $\endgroup$ – Arturo Magidin Jul 1 '12 at 20:56

$$ \lim_{\theta\to0} \frac{(\sin\theta)^2}{\theta} = \lim_{\theta\to0} \left(\sin\theta\frac{\sin\theta}{\theta}\right). $$ 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 $-\infty$.

  • $\begingroup$ +1. Importantly, this corrects @Kurt's placement of "1" inside his final limit expression. $\endgroup$ – Blue Jul 1 '12 at 19:49
  • $\begingroup$ @Michael Hardy, that clears things up. I can see now that limit will end up being 0. $\endgroup$ – Kurt Jul 1 '12 at 20:12
  • $\begingroup$ Even if the limits are to $\infty$, the product would be undefined as you can't do arithmetic with $\infty$. $\endgroup$ – Yves Daoust Sep 25 '19 at 14:57

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