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Evaluate the limit: $$ \lim_{x \to 0 } = x \cdot \sin\left(\frac{1}{x}\right) $$

So far I did:

$$ \lim_{x \to 0 } = x\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}\cdot x} $$

$$ \lim_{x \to 0 } = 1 \cdot \frac{x}{x} $$

$$ \lim_{x \to 0 } = 1 $$

Now of course I've looked around and I know I'm wrong, but I couldn't understand why. Can someone please show me how to evaluate this limit correctly? And tell me what I was doing was wrong.

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  • $\begingroup$ How did you go from the first line to the second? ($\lim_{x\rightarrow0}{\sin(1/x)\over 1/x}\ne 1$.) $\endgroup$ – David Mitra Dec 13 '14 at 16:37
  • $\begingroup$ I used: $\lim_{x \to 0} \frac{\sin(\frac{1}{x})}{\frac{1}{x}} = 1$ $\endgroup$ – FigureItOut Dec 13 '14 at 16:38
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    $\begingroup$ No, that's not right. Look closely... $\endgroup$ – David Mitra Dec 13 '14 at 16:39
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    $\begingroup$ If the argument of the $\sin$, call it $w$, tends to $0$, then ${\sin w\over w}$ tends to $0$. That's not what you have; the argument in your expression tends to $\pm\infty$, loosely speaking. On the other hand, $\lim\limits_{x\rightarrow\infty}{\sin(1/x)\over 1/x} =1$, since $1/x\rightarrow0$ as $x\rightarrow\infty$. $\endgroup$ – David Mitra Dec 13 '14 at 16:43
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    $\begingroup$ Just use the squeeze theorem: $$0\le|x\sin(1/x)|\le |x|\ \buildrel{x\rightarrow0}\over{\longrightarrow}\ 0.$$ $\endgroup$ – David Mitra Dec 13 '14 at 16:59
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Your proof is incorrect, cause you used incorrect transform, but it has already been stated. I'll describe way to solve it.

$$\lim_{x \to 0}\frac{\sin(\frac{1}{x})}{\frac{1}{x}} \neq 1$$

Hint: Solution is well know trick. Note $(\forall x \in \mathbb{R})\left(\sin(x) \in[-1;1]\right)$ (obvious) and use squeeze theorem to solve it.


Note simple implication.

$$ (\forall h \in \mathbb{R}) \left(\sin h \in [-1;1]\right) \Longrightarrow (\forall x,h \in \mathbb{R})(|x \cdot \sin h| \leq |x|)$$

So, true is inequality $|x \cdot \sin \frac{1}{x}| \leq |x|$, therefore (and because module is always non-negative) using squeeze theorem you receive limit.

$$\left(0 \leq\left | \lim_{x \to 0} x\cdot \sin \frac{1}{x} \right | \leq \lim_{x \to 0} \left| x \right| = 0 \right)\Longrightarrow \lim_{x \to 0}x \cdot \sin(x) = 0$$

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Hint: use the squeeze theorem.

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  • $\begingroup$ this still doesn't tell me why what I was doing wasn't legit.. I need to know so I won't so it again :S $\endgroup$ – FigureItOut Dec 13 '14 at 16:36
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    $\begingroup$ @user1326293 You mixed it up: $$\lim_{x\to 0}\frac{\sin(1/x)}{1/x}=\lim_{x\to \infty}\frac{\sin x}{x}\ne \lim_{x\to0}\frac{\sin x}{x}=1$$ $\endgroup$ – Vincenzo Oliva Dec 13 '14 at 16:40
  • $\begingroup$ You don't really need the squeeze theorem (depending on the formality of your course); just note that $x>>\sin(x)$ as $x$ gets large, so $\lim_{x\to\infty}\frac{\sin x}{x} = 0$. $\endgroup$ – apnorton Dec 13 '14 at 17:13
  • $\begingroup$ @anorton In the end, you're still using the fact that $-1\le \sin x \le 1$. And, you had to manipulate the limit, so that's less direct. $\endgroup$ – Vincenzo Oliva Dec 13 '14 at 17:20
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You cant do this because

$$\sin(1/x)$$ isnt defined. So you cant use the fact that

$$\lim_{x\to 0} \sin(\alpha)/\alpha = 1$$

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  • $\begingroup$ The function $f(x) = \sin(1/x)$ is certainly defined; what exactly do you mean? $\endgroup$ – apnorton Dec 13 '14 at 17:12
  • $\begingroup$ At $x=0$ it isnt. $\endgroup$ – Amad27 Dec 13 '14 at 17:13
  • $\begingroup$ Oh, ok. But more important than $f(x)$ not being defined at $x=0$, the limit to zero is also not defined. I see what you mean. $\endgroup$ – apnorton Dec 13 '14 at 17:14

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