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Can anyone help me to solve this problem?

Q: Show that $\cos(\frac{1}{x})$ cannot be continuously extended to $0$?

Notice: use this Theorem

Let $f:D\rightarrow \mathbb{R}$ and let $c\in D$. Then $f$ is continous at $c$ if and only if, whenever $X_n$ is a sequence in $D$ that converges to $c$, then $f(X_n)$ converges to $f(c)$.

I wish someone can solve this problem

Thanks

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    $\begingroup$ What have you tried so far? Can you find a sequence of values $t_n$ with $t_n\to\infty$ and $\cos(t_n)$ having no limit? Can you see how you might use that sequence to solve your problem? $\endgroup$ – Steven Stadnicki Apr 8 '13 at 6:55
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    $\begingroup$ Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help you if you show that you've tried the problem yourself. $\endgroup$ – Zev Chonoles Apr 8 '13 at 7:04
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Consider the sequence $X_n = \frac{1}{\pi n}$.

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  • $\begingroup$ Then supposing $f$ could be continuously extended, and taking $c=0$ in the theorem, what can be deduced about the value $f(0)$. $\endgroup$ – Jeppe Stig Nielsen Apr 8 '13 at 7:03
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Consider the sequences $$x_n = \dfrac1{2n \pi} \,\,\,\, \text{ and } x_n = \dfrac1{2 n \pi + \dfrac{\pi}2}$$ Both tend to zero. What happens to $\cos(1/x)$ along these sequences? (Recall that if a limit exists it has to be unique.)

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    $\begingroup$ thank you so much fried Marvis can you please explain more detail i need more explain thank you very much for helping $\endgroup$ – leena adam Apr 8 '13 at 6:57

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