Analyze if the limit for $x \in \mathbb{R}$ exists and calculate it if it does:

$$\lim_{x\rightarrow0, x>0}\sqrt{x}\cdot \sin\left(\frac{1}{x}\right)$$

For $x\rightarrow 0$:

$\sqrt{x}$ will converge to $0$ and $\sin\left(\frac{1}{x}\right)$ will diverge because it will always change its sign.

$\Rightarrow$ The complete term will diverge

For $x>0$:

$\sqrt{x}$ will diverge to $\infty$ and $\sin\left(\frac{1}{x}\right)$ will converge because $\lim_{x>0}\frac{1}{x}=0$ $\Rightarrow$ $\sin(0)=0$ and $\infty\cdot0=0$

$\Rightarrow$ The complete term will converge

Did I do everything correctly?

  • 2
    $\begingroup$ That's not what $$\lim_{x\to0,x>0}$$ means. Is means just "the limit as $x$ goes to zero, for positive $x$". $\endgroup$ – Arthur Jul 29 '16 at 8:32
  • $\begingroup$ So it means that x goes to positive zero? $\endgroup$ – cnmesr Jul 29 '16 at 8:38
  • $\begingroup$ If "positive zero" is something that even makes sense... But essentially, yes. $\endgroup$ – Arthur Jul 29 '16 at 8:42
  • $\begingroup$ It's not true that $\lim_{x \to 0, x>0}\frac{1}{x}=0$. $\endgroup$ – Olivier Oloa Jul 29 '16 at 8:45
  • $\begingroup$ Olivier, what is it then? I think $\lim_{x\rightarrow0, x>0}\frac{1}{x}=+\infty$ Correct? $\endgroup$ – cnmesr Jul 29 '16 at 8:49

No, what you wrote is not correct.

A possible route is to just observe that $$ \left|\sqrt{x}\cdot \sin\left(\frac{1}{x}\right)\right|\le \sqrt{x} $$ then letting $x \to 0^+$.

  • $\begingroup$ @cnmesr We don't have $\lim_{x\rightarrow0, x>0}\sin\left( {\frac{1}{x}}\right )=0$, this is wrong. $\endgroup$ – Olivier Oloa Jul 29 '16 at 8:55
  • $\begingroup$ $$\lim_{x\rightarrow0, x>0}\sqrt{x} \cdot \lim_{x\rightarrow0, x>0}sin\left( {\frac{1}{x}}\right )=0\cdot sin \left(\infty \right )=0$$ $\endgroup$ – cnmesr Jul 29 '16 at 8:57
  • $\begingroup$ It's better to write $\left|\lim_{x\rightarrow0, x>0}\left(\sqrt{x} \cdot\sin\left( {\frac{1}{x}}\right )\right)\right|\le \lim_{x\rightarrow0, x>0}\sqrt{x}=0.$ $\endgroup$ – Olivier Oloa Jul 29 '16 at 9:00
  • $\begingroup$ I think limit no exist because sinus of infinity has different signs, this multiplied by small value (almost zero) will still have different signs, thats why no limit, right? $\endgroup$ – cnmesr Jul 29 '16 at 9:01
  • $\begingroup$ Yes $\sin (1/x)$ is wildly oscillating as $x \to 0$, but it is absoulutely bounded by 1... $\endgroup$ – Olivier Oloa Jul 29 '16 at 9:03

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