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.

  • $\begingroup$ 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
  • 2
    $\begingroup$ No, that's not right. Look closely... $\endgroup$ – David Mitra Dec 13 '14 at 16:39
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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.

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


Hint: use the squeeze theorem.

  • $\begingroup$ 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
  • 7
    $\begingroup$ @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

You cant do this because

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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.