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To solve $ \lim_{x \to 0} (\frac{x\cdot\sin{x}}{|x|}) $ I have separated the function in two limits:

\begin{align} \lim_{x \to 0} (\frac{x\cdot\sin{x}}{|x|}) = \lim_{x \to 0} \frac{x}{|x|} \cdot \lim_{x \to 0} \sin{x} \end{align}

Now solving independently:

\begin{align} \lim_{x \to 0} \frac{x}{|x|} = \lim_{x \to 0} \text{sign}(x) \, \, \, \text{does not exist} \end{align}

And:

\begin{align} \lim_{x \to 0} \sin{x} = 0 \end{align}

I've checked graphically and I'm aware that the limit is $ 0 $, however, I'm not sure that multiplying "undefined" by $ 0 $ is something allowed.

Can I do this? If not, what could be another way of solving the limit?

Edit:

Just realized I can split the function in a different way, knowing that $ \lim_{x \to 0} \frac{\sin{x}}{|x|} = 1 $

\begin{align} \lim_{x \to 0} (\frac{x\cdot\sin{x}}{|x|}) = \lim_{x \to 0} (\frac{|x|\cdot\sin{x}}{x}) = \lim_{x \to 0} |x| \cdot \lim_{x \to 0} \frac{\sin{x}}{x} = 0 \cdot 1 = 0 \end{align}

How about this second approach?

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  • $\begingroup$ Note that $\lim_{x \to a}(f \cdot g)=(\lim_{x \to a} f) \cdot (\lim_{x \to a}g)$ requires the existence of both limits on the right side which is not the case here. Instead you can say as $x \to 0^{+}$, then the function is $\sin x$ and as $x \to 0^{-}$, the function is $-\sin x$. $\endgroup$
    – Anurag A
    Aug 7, 2020 at 19:00
  • $\begingroup$ Why on the right side specifically? Shouldn't be, in general, the existence of both limits? $\endgroup$
    – Jon
    Aug 7, 2020 at 19:01
  • $\begingroup$ $$\lim_{x \to 0} f(x) = 0 \iff \lim_{x \to 0} \lvert f(x)\rvert = 0$$ $\endgroup$ Aug 7, 2020 at 19:02
  • $\begingroup$ We have that $\lim_{x \to 0} \frac{\sin{x}}{|x|} = \pm 1$. $\endgroup$
    – user
    Aug 7, 2020 at 19:11
  • $\begingroup$ Thanks, @user. Made a typo with the $ |x| $ in the denominator, should be just $ x $ $\endgroup$
    – Jon
    Aug 7, 2020 at 19:13

5 Answers 5

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$$\lim_{x\to 0} \frac{ x\cdot \sin x}{|x|} = \lim_{x\to 0} \frac{ |x| \cdot \sin x}{x} = \lim_{x\to0} |x| \cdot \lim_{x\to0}\frac{\sin x}{x} = 0\cdot 1 = 0.$$

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  • $\begingroup$ Very clever method! $\endgroup$
    – user
    Aug 7, 2020 at 19:31
  • $\begingroup$ insane! wouldn't have thought! $\endgroup$
    – anonymous
    Aug 18, 2021 at 22:25
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You can separate a limit into two only if both exist. Here as you point out, one of them does not, so this step is wrong.

You can say that $\operatorname{sgn}(x)$ is bounded, $\sin(x)$ goes to $0$, so their multiplication goes to $0$ as well.

Alternatively, calculate $x\to 0^+$ and $x\to 0^-$ and see they are equal.

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Approach $0$ from the right:

$$\lim_{x \to 0^+} \frac{x\sin{x}}{|x|}=\lim_{x \to 0^+}\frac{x\sin{x}}{x}=\lim_{x \to 0^+}\sin x=0.$$

Then approach $0$ from the left:

$$\lim_{x \to 0^-} \frac{x\sin{x}}{|x|}=\lim_{x \to 0^-}\frac{x\sin{x}}{-x}=\lim_{x \to 0^-}-\sin x=0.$$

Since $\lim_{x \to 0^+}\frac{x\sin{x}}{|x|}=0=\lim_{x \to 0^-}\frac{x\sin{x}}{|x|}$, the limit is zero.

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The important fact is that $\frac{x}{|x|}$ is bounded that is

$$-1\le\frac{x}{|x|}\le 1$$

therefore

$$0\le \left|\frac{x\cdot\sin{x}}{|x|}\right|\le|\sin x| \to 0$$

and we can conclude by squeeze theorem.

Alternatively we can use that

$$\frac{|\sin{x}|}{|x|} \to 1$$

and therefore

$$0\le \left|\frac{x\cdot\sin{x}}{|x|}\right|=|x|\frac{|\sin{x}|}{|x|} \to 0\cdot 1=0$$

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  • $\begingroup$ Thanks for your answer. I knew about the approach using the squeeze theorem but I'm afraid I don't understand how to use in this context. I'm aware that the signum function is bounded between -1 and 1, and sin(x) is also bounded in the same way. With that information, how can I discover that my original function is between them? $\endgroup$
    – Jon
    Aug 7, 2020 at 19:23
  • $\begingroup$ @Jon We know that $x/|x|$ is bounded whereas $\sin x \to 0$, for that reason the product goes to zero. To formalize that we invoque squeeze theorem but it is a very intuitive fact. $\endgroup$
    – user
    Aug 7, 2020 at 19:26
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Let $f(x)=\frac{x\sin(x)}{\left| x\right|}$. If $x$ approaches $0$ from the right, $x>0$, so $\left| x\right| = x$. Hence, $$\lim_{x\rightarrow 0^+} f(x)$$$$=\lim_{x\rightarrow 0^+}\frac{x\sin(x)}{x}$$$$=\lim_{x\rightarrow 0^+}\sin(x)$$$$=0.$$ Because $f(x)=f(-x)$, $$\lim_{x\rightarrow 0^-}f(x)=\lim_{x\rightarrow 0^+}f(x)=0.$$ Therefore, $\lim_{x\rightarrow 0}f(x)=0.$

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