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My Teacher wants us to argue that this equation below is true. I am just curious how I would actually explain this? I mean I understand how its true but how can I argue this?

$$\lim_{\theta\rightarrow 0^+}\frac{\cos\frac{1}{\theta}\sin\theta-\theta\cos\frac{1}{\theta}}{\theta} = 0$$

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There is no equation here. – William Jun 24 '12 at 23:53
What do you want to argue? Do you want to argue that $$\lim_{\theta \to 0^+} \dfrac{\cos(1/\theta) \sin(\theta) - \theta \cos(1/\theta)}{\theta} = 0?$$ – user17762 Jun 24 '12 at 23:55
Thats what the problem asked which i dont understand – soniccool Jun 25 '12 at 0:59
If the problem is to show that the limit is zero, can you please edit that into the question? As it stands, you haven't given an equation, so you haven't asked a question. – Gerry Myerson Jun 25 '12 at 1:44
@soniccool: I don't understand "u understand how its true": how do you understand that? Also, explaining a little of what you know about limits, and what you've tried already, would be useful. – Ben Millwood Jun 25 '12 at 3:03

$$ \lim_{\theta \rightarrow 0^+} \cos \left ( 1 \over \theta \right ) \left ( \sin \theta - \theta \over \theta \right) $$ Let, $ \frac{1}{\theta} = \phi $, then we have for all real values of $ \phi $ $$ -1 \leq \left ( \cos \phi \right ) \leq 1 $$

EDIT:: taking comments into consideration $$ - \left| \dfrac{\sin \theta}{\theta} - 1 \right| \le \cos(1/\theta) \left(\dfrac{\sin \theta}{\theta} - 1 \right) \le \left|\dfrac{\sin \theta}{\theta} - 1 \right| $$ $$ -1 \times 0 \leq \lim_{\phi \rightarrow \infty } \cos \phi \times \lim_{\theta \rightarrow 0 } \left ( \frac{\sin \theta} \theta - 1 \right ) \leq 1 \times 0 $$ $$ \text{Or, using squeeze theorem, we have, } \lim_{\theta \rightarrow 0^+} \cos \left ( 1 \over \theta \right ) \left ( \sin \theta - \theta \over \theta \right) = 0 $$

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I think I get what you're saying here, but this answer isn't really clear as it is. What do you mean by some value? Where does the zero come from? Why isn't it a problem that we divide by $\theta$ and then take the limit as $\theta$ tends to 0? (As a matter of fact I know the answer to these questions. But they're really the substance of the question, and need addressing, IMO) – Ben Millwood Jun 25 '12 at 3:02
Although this answer is essentially correct, I would have been happier to see an explanation along the following lines: If $g$ is a bounded function, say $-M<g(x)<M$ for all $x$ in the domain of $g$, and if $f$ is a function whose limit at $a$ is zero, $a$ being an element of the closure of the domain of $g$, then $\lim_{x\to a}f(x)g(x) = 0$. What’s going on between the parentheses in $\cos(blah)$ is immaterial: that was a red herring. – Lubin Jun 25 '12 at 3:15
The ironic thing is that sometimes, if a high rep user does this, he/she gets many upvotes, but we tend to nit pick when it is another person. – Pedro Tamaroff Jun 25 '12 at 3:24
Dear experimentX, As written, your final line is not correct, since $\lim_{\phi \to \infty} \cos \phi$ does not exist. In any case, as @Lubin notes, manipulating the argument inside the $\cos$ function is irrelevant. Regards, – Matt E Jun 25 '12 at 3:41
How would you write it?? the best i can think of it now is $$ -1 \times 0 \leq \lim_{\phi \rightarrow \infty } \cos \phi \times \lim_{\theta \rightarrow 0 } \left ( \frac{\sin \theta} \theta - 1 \right ) \leq 1 \times 0 = 0 $$ – Santosh Linkha Jun 25 '12 at 3:46

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