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Alright, scratch everything below the line. Let me present one cohesive question not marred by repeated edits.

The limit $\lim_{x\to a}f(x)=L$ exists iff for every $\epsilon>0$ there is a $\delta>0$ such that $|f(x)-L|<\epsilon$ when $0<|x-a|<\delta$.

Thus, $\lim_{x\to0}\sin\left(\frac1x\right)$ does not exist because, being that it oscillates infinitely near $0$, there is no $\epsilon,\delta$.

On the other hand, with the limit$$\lim_{x\to0}\frac{\sin\left(x\sin\left(\frac1x\right)\right)}{x\sin\left(\frac1x\right)}\\\lim_{x\to0}x\sin\left(\frac1x\right)=0\\y=x\sin\left(\frac1x\right)\\\lim_{y\to0}\frac{\sin y}{y}=1\\\lim_{x\to0}\frac{\sin\left(x\sin\left(\frac1x\right)\right)}{x\sin\left(\frac1x\right)}=1$$

this proof can be shown. However, since $\sin\left(\frac1x\right)$ oscillates infinitely, by the same definition of limit we used to show the above, the limit does not exist. How do I resolve this discrepancy?

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    $\begingroup$ You probably need to say what is your definition for a limit $\lim_{x\to 0} f(x)$ where every neighborhood of zero contains points where $f(x)$ is undefined. $\endgroup$ – GEdgar Apr 17 '13 at 16:13
  • $\begingroup$ I don't understand what you're trying to say... I'm just using the definition of limits in my textbook, the same one that tells me $\lim_{x\to0}\sin\frac1x$ is undefined, or $\lim_{x\to0}\frac{\sin x}{x}=1$. Except here I have a mix of the two. Shouldn't it be consistent? $\endgroup$ – user72273 Apr 17 '13 at 16:20
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    $\begingroup$ The point is, $\lim \sin \frac{1}{x}$ is not defined, but $\lim x \sin \frac{1}{x}$ exists, and you only want to look at the latter. Also: extensive edits make the four answers that have appeared look a little strange (it might have been better to keep on adding edits regardless). $\endgroup$ – Jakub Konieczny Apr 17 '13 at 16:47
  • $\begingroup$ My comment is: $x \sin(1/x)$ vanishes in some points in every neighborhood of zero, so something with this in the denominator is undefined at those points. $\endgroup$ – GEdgar Apr 17 '13 at 20:04
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A useful thing to know about limits, is that if you are evaluating limit of the form $\lim_{x \to a} f(g(x))$, and you know that $\lim_{x \to a} g(x) = b$ then $\lim_{x \to a} f(g(x)) = \lim_{y \to b} f(y)$ (assuming everything is nice and continuous). So, because you have already worked out the limit $\lim_{x\to 0} x \sin\frac{1}{x} = 0$, you get: $$ \lim_{x \to 0} \frac{\sin(x \sin\frac{1}{x})}{x \sin\frac{1}{x}} = \lim_{y\to 0}\frac{\sin y}{y} = 1$$

Edit

Let me be a little more precise. Suppose you know that $\lim_{y \to b} f(y) = L$ (so for $\varepsilon$ you have $\delta_f(\varepsilon)$ such that if $|y -b| < \delta_f(\varepsilon)$ you have $|f(y) - L| < \varepsilon$) and that $\lim_{x \to a} g(x) = b$ (so for $\varepsilon$ you have $\delta_g(\varepsilon)$ such that if $|x -a| < \delta_g(\varepsilon)$ you have $|g(x) - b| < \varepsilon$). I claim that then, with no extra assumption $\lim_{x \to a} f(g(x)) = L$. With $g(x) = x \sin\frac{1}{x}$ and $f(y) = \frac{1}{y}\sin y$, this solves your problem.

The reasoning is as follows: Take $\varepsilon > 0$, and define $\delta := \delta_g(\delta_f)$. I claim that if $|x-a| < \delta$ then $|f(g(x))-L| < \varepsilon$. First, because $|x-a| < \delta_g(\delta_f(\varepsilon))$, you have $|g(x)-b| < \delta_f(\varepsilon)$. Next, because $|g(x)-b| < \delta_f(\varepsilon)$, you have $|f(g(x)) - L| < \varepsilon$, as promised.

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    $\begingroup$ I think there's a problem: $f(y)$ is not defined when $y=0$, while $x=\frac{1}{n\pi}$ gives rise to $y=g(x)=0$. $\endgroup$ – Shane O Rourke Apr 17 '13 at 17:12
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The limit is undefined: no matter how small a $\delta>0$ you choose, there will exist positive $n\in\mathbb{Z}$ with $0<\frac{1}{n\pi}<\delta$. Let's call your function $f(x)$. Then the denominator of $f(x)$ is zero, making $f(x)$ itself undefined at such values. Therefore it would be untrue to say "$|f(x)-l|<\epsilon$ for all $0<|x|<\delta$", since $f(x)$ is undefined for $x=\frac{1}{n\pi}$.

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    $\begingroup$ I think you are correct, and I missed this when writing the answer. It could be said that the function whose limit is being computed is not well defined at the points we need. I'd say that should be added in the problem formulation somehow (like, specify that the limit is taken along points where the function is specified, or adding something like "$\frac{\sin x}{x} := 1$ for $x = 0$"). $\endgroup$ – Jakub Konieczny Apr 17 '13 at 17:48
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    $\begingroup$ The fundamental problem here is that the theorem "if $\lim_{x \to a}g(x) = b$ and $\lim_{x \to b}f(x) = L$ then $\lim_{x \to a}f\{g(x)\} = L$ is false. To make it true requires an extra condition that $g(x) \neq b$ in the deleted neighborhood of $x = a$. Its bit unfortunate that a technically wrong answer has been accepted. But I think many books have missed this point and it is a very subtle error and difficult to detect otherwise. $\endgroup$ – Paramanand Singh Nov 26 '13 at 18:09
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As we know that $\lim_{x \to 0} \frac{\sin x}{x} = 1$. Point to be note here is that $lim_{x\to 0} x = 0$. I hope you got your answer with this hint.

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(Hint) What is $\displaystyle \lim_{x \to 0} \frac{\sin x}{x}$?

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  • $\begingroup$ I already considered this, can you look at my edit? $\endgroup$ – user72273 Apr 17 '13 at 16:12
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You have something of the form $\frac{\sin(u)}u$, if $u=x\sin(1/x)$. Can you transform this into a limit of $u$? If $x\to 0$, $u\to?$

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I'll prove this result: if $\displaystyle\lim_{x\to a} g(x)=b$ then $\displaystyle\lim_{x\to a} f(g(x))=\ell=\lim_{y\to b}f(y)$.

Let $\epsilon>0$, there's $\delta>0$: if $|x-a|<\delta|$ then $|f(g(x))-\ell|<\epsilon$

For the founded $\delta$ there's $\delta'>0$ and if $|x-a|<\delta'$ then $|g(x)-b|<\delta$

hence take $y=g(x)$ and if $|x-a|<\min(\delta,\delta')$ then $|y-b|<\delta$ and $|f(y)-\ell|<\epsilon$.

From this I think you can apply the result to your question.

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    $\begingroup$ you need to add extra hypothesis that $g(x) \neq b$ in deleted neighborhood of $a$, otherwise the result is false. $\endgroup$ – Paramanand Singh Nov 26 '13 at 18:10

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