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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.

Martin Sleziak
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FigureItOut
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  • How did you go from the first line to the second? ($\lim_{x\rightarrow0}{\sin(1/x)\over 1/x}\ne 1$.) – David Mitra Dec 13 '14 at 16:37
  • I used: $\lim_{x \to 0} \frac{\sin(\frac{1}{x})}{\frac{1}{x}} = 1$ – FigureItOut Dec 13 '14 at 16:38
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    No, that's not right. Look closely... – David Mitra Dec 13 '14 at 16:39
  • Can you please tell me :D? – FigureItOut Dec 13 '14 at 16:40
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    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 $x\rightarrow\infty$. – David Mitra Dec 13 '14 at 16:43
  • OK, let's say I got you.. so you say I can use $$\lim_{x \to c} \frac{sinc}{c} = 1$$ only if $c$ is getting closer to $0$? and because $\frac{1}{x}$ approaches $\infty$ when $ x \to 0$ that is not valid? Do you have an idea how to evaluate it using the trigonometric identities? – FigureItOut Dec 13 '14 at 16:53
  • Almost right; your notation is off. If $\lim_{x\rightarrow c} g(x)=0$, then $\lim_{x\rightarrow c} {\sin g(x)\over g(x)}=0$. You can't appeal to this for $g(x)=1/x$ and $c=0$. – David Mitra Dec 13 '14 at 16:57
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    Just use the squeeze theorem: $$0\le|x\sin(1/x)|\le |x|\ \buildrel{x\rightarrow0}\over{\longrightarrow}\ 0.$$ – David Mitra Dec 13 '14 at 16:59
  • Related post: http://math.stackexchange.com/questions/891204/solve-lim-x-to-0x-sin-frac1x-using-lim-theta-to-0-frac-sin-t Found through catalog of limits. – Martin Sleziak Nov 19 '15 at 14:34
  • Hint: use the squeeze theorem. – Vincenzo Oliva Dec 13 '14 at 16:35

2 Answers2

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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-known 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.

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

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.

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

Cheese Cake
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Tacet
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You cant do this because

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

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

Amad27
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