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$\displaystyle\lim_{\theta \to 0} \theta \sin\left(\frac{1}{\theta}\right)$

plugging in 0, I get $0*\sin(\frac{1}{0})$ would this be an indeterminant form? Or would this be acceptable to being the limit = 0?

If it's indeterminant, how would I reduce and retry?

When I graph, I get a horizontal line at 0, which makes me think my initial assumption was correct. I'm just a little troubled with the idea of sin(1/0)...

Thanks for the feedback and input.

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HINT: $|\sin (x)| \leq 1$ for all real numbers $x$. –  JavaMan Jun 20 '11 at 19:19
SQUEEZE THEOREM.... I should have known! Thanks. –  OghmaOsiris Jun 20 '11 at 19:24
You should not have a horizontal line near $0$ –  Henry Jun 20 '11 at 21:50
I'm guessing my graphing calculator didn't know how to graph the function, as that's what my calculator was giving me, lol. –  OghmaOsiris Jun 20 '11 at 21:55
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3 Answers 3

up vote 1 down vote accepted

You may want to read this first.

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Take a close look at your graph near $0$. For my comfort I will change the variable from $\theta$ to $x$. You will observe that $x\sin(1/x)$ wiggles desperately, but with decreasing amplitude, so the limit appears to be $0$. For a proof that the graphing calculator or program is not misleading you (they sometimes do), the hint given by Shai Covo should be enough.

The main reason for my post is to warn you against the reasoning that in this case happened to produce the right answer.

Look in particular at $x \times \frac{1}{\sin x}$, as $x$ approaches $0$. If you put $x=0$ as you seemed to in your reasoning, you may think that you should get an answer of $0$. However, $$\lim_{x \to 0} \left(x \times \frac{1}{\sin x}\right)=1$$

You may already have seen an argument for this, or probably for the close relative $(\sin x)/x$, when the derivative of $\sin x$ was first calculated. Informally, you can do a partial verification by graphing $\frac{x}{\sin x}$ (make sure that your calculator is in radian mode).

The same type of reasoning that happened to produce the right answer in one case produces the wrong answer in another! Thus this kind of reasoning is not to be trusted. We need to use tools that we can trust.

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Hint: $|\theta \sin(1/\theta)| \leq |\theta|$ for all $\theta \neq 0$.

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When considering the limit, you should only be concerned with $\theta$ such that $0 < |\theta| < \delta$. –  Shai Covo Jun 20 '11 at 19:24
The function can be defined arbitrarily at $0$; this does not affect the limit. –  Shai Covo Jun 20 '11 at 19:25
If you want the function, call it $f$, to be continuous at $0$, then define $f(0)=0$. –  Shai Covo Jun 20 '11 at 19:29
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