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How do I evaluate $$\lim_{\theta \to 0^+}\frac{\sin\theta}{\theta^2}?$$

I tried the following:

$$\lim_{\theta \to 0^+}\frac{\sin\theta}{\theta^2} = \lim_{\theta \to 0^+}\frac{1}{\theta}\cdot \lim_{\theta \to 0^+}\frac{\sin\theta}{\theta} = \lim_{\theta \to 0^+}\frac{1}{\theta} = +\infty$$

However, I feel that there is an error with my work, since I believe it isn't acceptable to separate a limit when it separates into something that has a value of infinity. Is there an issue with my work here?

Thomas
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okarin
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3 Answers3

14

Right, it's not necessarily valid to separate the limit like you did; but it does at least suggest the right technique: Since $$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$$

It's true that $\sin \theta / \theta > 1/2$ for all $\theta$ sufficiently small. As a result,

$$\frac{\sin \theta}{\theta^2} = \frac{\sin \theta}{\theta} \cdot \frac 1 {\theta} > \frac 1 {2 \theta}$$

Since the right side tends to infinity as $\theta \to 0^+$, the left hand side does too.

4

You may try in such a way: $\forall \epsilon >0$ we have $\lim_{\theta\rightarrow 0^{+}} \frac{sin(\theta)}{\theta^2} \ge \lim_{\theta\rightarrow 0^{+}} \frac{sin(\theta)}{\theta\epsilon} = \frac{1}{\epsilon}$. since we know limit $\ge \frac{1}{\epsilon}$ for all $\epsilon$, we can conclude that $\lim_{\theta\rightarrow 0^{+}} \frac{sin(\theta)}{\theta^2} = \infty$ or simply say the limit does not exist.

Alex Reed
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T. Bongers already gave the best answer, but a slightly different way to do it is to note that the limit is $$ \lim_{\theta \to 0^+} \frac{\sin(\theta)}{\theta}\frac{1}{\theta}. $$ You, probably, already know that $$ \lim_{\theta \to 0}\frac{\sin(\theta)}{\theta} = 1. $$ So, you are multiplying something that is approaching $1$ by something that is approaching infinity. The whole thing is going to grow without bound.

Thomas
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  • This description is equivalent to separating the limits, which as the OP noted, you cannot do if either limit does not exist. – Euro Micelli May 01 '14 at 06:15
  • @EuroMicelli: I didn't write the limit as the product of two limits. Of course one would have to know that if you have a product where one of the factors is bounded while the other one grows without bound, then the product grows without bound. (Hardly worth a downvote ...) – Thomas May 01 '14 at 12:08
  • Which is why I didn't vote it down. If you were down-voted, it was someone else. I'm still not convinced the argument is valid, but I'm not sure enough to 'vote it' one way or another. – Euro Micelli May 01 '14 at 13:27
  • @EuroMicelli: Alright. In that case I request the downvoter tell me why the downvote... – Thomas May 01 '14 at 13:29