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Given function is $f(x)= x|\cos(2\pi/x)|$ for all $x$ leaving zero. and $f(x)= 0$ at $x= 0$ ,how can we able to evaluate the right and left hand limit of $x= 0$,as such here values are drastically changing in terms of magnitude and sign wise near $x= 0$ ..

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    $\begingroup$ Why not use the fact that $0 \leq |f(x)| \leq |x|$? $\endgroup$ May 16, 2021 at 11:44
  • $\begingroup$ But $|\cos|$ is still bounded in $[0,1]$, right? What's zero multiplied by a bounded quantity... $\endgroup$ May 16, 2021 at 11:47
  • $\begingroup$ Hmm i agree about the idea of applying sandwich theorem , i am just referring to without using that idea $\endgroup$
    – Orion_Pax
    May 16, 2021 at 11:54
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    $\begingroup$ @ShubhamJohri Zero multiplied by any quantity is zero. Perhaps you meant what is something tending to zero times something bounded...? $\endgroup$
    – DonAntonio
    May 16, 2021 at 11:59
  • $\begingroup$ Yes, please excuse the inaccuracy $\endgroup$ May 16, 2021 at 12:04

3 Answers 3

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Whenever you encounter a question about limits, you should first check which kind of indeterminate form that function is forming. Many questions look confusing can be solved by considering this.

Consider the function given in your question. If you put the limits in to check which indeterminate form it is, you will see that it is not in any indeterminate form. So you have to show the continuity using some arguments.

You will get $f(0^+)= 0 \times |\cos(2π/0)|$. What is $\cos(2π/0)$? It tends to $\cos(∞)$, which oscillates between $0$ and $1$. So $f(0^+)$ is $0$ times a number between 0 and 1, which is actually zero.

You can use similar argument to show for Left hand limit, i.e. $f(0^-)$.

So LHL$=$RHL$=f(0)=0$, hence, the function given is continuous.

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$$f(x)= x|\cos(2\pi/x)|$$

Right hand limit:

$$\lim_{x\rightarrow 0^+} f(x)= 0^+\times|\cos(2\pi/0^+)|=0^+\times|\cos(+\infty)|$$

Since $\cos(x)$ is bounded: $$0 \leq |\cos(+\infty)| \leq 1$$

We can see that because $0^+$ multiplies a bounded value is $0$, the limit is $0$.
Or, multiply the inequality with $0^+$:

$$0^+ \times 0 \leq 0^+ \times|\cos(+\infty)| \leq 0^+ \times 1$$ $$0 \leq 0^+ \times|\cos(+\infty)| \leq 0$$

Therefore, $$\lim_{x\rightarrow 0^+} f(x)=0^+\times|\cos(+\infty)|=0$$

Left hand limit:

$$\lim_{x\rightarrow 0^-} f(x)= 0^-\times|\cos(2\pi/0^-)|=0^+\times|\cos(-\infty)|$$

$$0 \leq |\cos(-\infty)| \leq 1$$

Similarly, $0^- \times|\cos(-\infty)|$ is also $0$. Multiply the inequality with $0^-$:

$$0^- \times 0 \geq 0^- \times|\cos(-\infty)| \geq 0^- \times 1$$ $$0 \geq 0^- \times|\cos(-\infty)| \geq 0$$

Therefore,

$$\lim_{x\rightarrow 0^-} f(x)=0^-\times|\cos(-\infty)|=0.$$

Finally,

$$\lim_{x\rightarrow 0^-} f(x)= \lim_{x\rightarrow 0^+} f(x)= f(0)=0.$$ The function is continuous.

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  • $\begingroup$ Thanks understood $\endgroup$
    – Orion_Pax
    Dec 17, 2021 at 7:29
  • $\begingroup$ @Orion_Pax Could you tell what was missing in my answer? It will be helpful for me, to write a better answer from next time :) $\endgroup$
    – Albedo
    Jan 28 at 8:46
  • $\begingroup$ Nothing it was accurate i was actually confused whom to give between you and other one $\endgroup$
    – Orion_Pax
    Jan 28 at 9:32
  • $\begingroup$ I could have given to both if it was there $\endgroup$
    – Orion_Pax
    Jan 28 at 9:32
  • $\begingroup$ For now i can give tick as i already gave my bounty to him . But yeah may this be treated as two parts of a prize $\endgroup$
    – Orion_Pax
    Jan 28 at 9:33
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Theorem. Let $f(x)$ a $\tau-$period function, different from constant one. Then, it has no limit when $x\to+\infty$ or $x\to -\infty$.

Using only this theorem, one can think he is not able to solve the $\lim_{x\to 0}x\cdot\left|\cos\left(\frac{2\pi}{x}\right)\right|$. But, here it comes another theorem.

Theorem. Let $f(x)$ a bounded function in $I(x_0)$ and $g(x)$ such that $\lim_{x\to x_0}g(x)=0$, than: $$\lim_{x\to x_0}f(x)\cdot g(x)=0$$ Here, let $f(x)=\left|\cos\left(\frac{2\pi}{x}\right)\right|$ that is bounded in $[-1,1]$. Let $g(x)=x$. By the previous theorem, we can say that: $$\lim_{x\to 0}x\cdot\left|\cos\left(\frac{2\pi}{x}\right)\right|=0$$ You can extend the function with continuity, letting: $$\tilde h(x)=\left\{\begin{matrix} x\cdot\left|\cos\left(\frac{2\pi}{x}\right)\right| \;\;x\neq 0\\ 0\;\;\;\;\,\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x = 0 & \end{matrix}\right.$$

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  • $\begingroup$ Thank u Sir understood $\endgroup$
    – Orion_Pax
    Dec 17, 2021 at 7:30

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