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By Cauchy's criterion of limit (not sequencial criterion), show that $$\lim_{x\to 0}(\sin{\frac{1}{x}}+x\cos{\frac{1}{x}})$$ does not exist.

Cauchy's criterion of limit

$\lim_{x\to c}f(x)=l$ iff for every $\epsilon>0$, there exists $\delta$ such that $$|f(x_2)-f(x_1)|<\epsilon$$ for $0<|x_1-c|<\delta$ and $0<|x_2-c|<\delta$.

Please suggest $x_2, x_1$ and help me to solve the problem.

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One part of Cauchy's Criterion says that

RESULT: If $\exists \epsilon>0$ such that $\forall \delta >0$, we can find $x_1,x_2$ satisfying $0<|x_1-a|<\delta$ and $0<|x_2-a|<\delta$ but $|f(x_2)-f(x_1)|\geq \epsilon$ then $\lim_{x\to a}f(x)$ does not exist.

Let us write $$f(x)=\sin\frac{1}{x}+x\cos\frac{1}{x}.$$

Take $\epsilon\leq 2$. Let $\delta>0$. By Archimedean property,we can find $n\in\Bbb N$ such that $\frac{1}{n}<\delta$. Take $$x_1=\frac{1}{\frac{3\pi}{2}+2\pi n}\qquad\text{and}\qquad x_2=\frac{1}{\frac{\pi}{2}+2\pi n}.$$ Notice that $0<x_1<\frac{1}{n}<\delta$ and $0<x_2<\frac{1}{n}<\delta$. Notice that $\sin x_2=1$, $\sin x_1=-1$, $\cos x_2=0$, and $\cos x_1=0$. With this, we get $$|f(x_2)-f(x_1)|=|1-(-1)|=2\geq\epsilon.$$
Apply the result and we are done.

NOTE: That was the idea behind the hint given by @5xum. The only difference is that our delta and epsilon were interchanged. Since the OP want for clarification, I rather post an answer than to put all of these in the comments.

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  • $\begingroup$ Thanks a lot. Yes it a good idea. Instead of $\lim_{x\to 0}(\sin{\frac{1}{x}}+x\cos{\frac{1}{x}})$ if the limit is like $\lim_{x\to 0}(\sin{\frac{1}{x}}+x\sin{\frac{1}{x}})$ then your do work or not? $\endgroup$ – user1942348 Mar 13 '17 at 14:57
  • $\begingroup$ @user1942348 Well, its another problem and maybe the approach is different to the previous one. Try to solve it first by using the approach that we had and see what happens. That is a good exercise to you. $\endgroup$ – Juniven Mar 13 '17 at 15:11
  • $\begingroup$ I can solve it by my approach, but your method gives me trouble. Would you suggest something $\endgroup$ – user1942348 Mar 13 '17 at 15:14
  • $\begingroup$ @user1942348 Well, that is also an interesting question. As of now, I don't know what to do--) $\endgroup$ – Juniven Mar 13 '17 at 15:23
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Hint:

You can find, for any $\epsilon > 0$, a value of $x_1$ such that $0<x_1<\epsilon$ and $\sin\frac1{x_1}=1$, and a value of $x_2$ such that $0<x_2<\epsilon$ and $\sin\frac{1}{x_2}=-1$.

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  • $\begingroup$ $x_2=\frac{1}{(4n-1)\pi/2}$ and $x_1=\frac{1}{(4n+1)\pi/2}$ then $|\sin\frac1{x_2}-\sin\frac1{x_1}|=2$ not less than $\epsilon$. $\lim \sin\frac1{x}$ does not exist. Please what to do next? $\endgroup$ – user1942348 Mar 13 '17 at 11:56
  • $\begingroup$ @user1942348 well if $|x|<\epsilon$, then $|x\cos x| = |x||\cos x| \leq |x|\cdot 1 \leq \epsilon$. $\endgroup$ – 5xum Mar 13 '17 at 11:58
  • $\begingroup$ One lim does not exist and other lim exists, so lim of their sum does exists? $\endgroup$ – user1942348 Mar 13 '17 at 12:09
  • $\begingroup$ @user1942348 No, that's not correct. You already have the values of $x_1,x_2$. Now just look at your entire function (not just the $\sin$ part)... $\endgroup$ – 5xum Mar 13 '17 at 12:10
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    $\begingroup$ It should be $x=\frac{3\pi}{2}+2\pi n$ where $n=0,1,2,3,\dots$ $\endgroup$ – Juniven Mar 13 '17 at 13:15
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Note first that $$\lim_{x \to 0}x \cos \frac 1x=0$$ so you can safely focus on $\sin \frac 1x$. Now consider $x_n=\frac 1{2n\pi+\frac{\pi}2}$ and $y_n=\frac1{n\pi}$ for $n\to \infty$ and see what happens.

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