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Let $f:\Bbb R\rightarrow \Bbb R\;$ s.t. $f(x) =x\sin{1\over x}$

Q1 : Is the $f$ continuous at $x=0$ ? If so, what is the value of $f(x)$ at $x =0$ ?


My Solution : Since $\sin{1\over x}$ trapped in a range of $-1 \text{ to } 1$, when $x$ goes to 0, $\lim\limits_{x\to 0} x\sin{1\over x} = 0$. Thus it's continuous and the value is $0$.


Q2: To check the Local linearity at $x =0$, we need to check whethere there exists derivative of $x\sin{1\over x}$ at $x =0$.

Firstly, we need to check whether $\lim\limits_{h\to0+}{f(x+h)-f(x)\over h} = \lim\limits_{h\to0-}{f(x+h)-f(x)\over h} $ when $x=0$

So by applying $f(0) = 0$ to both sides, we get:

$\lim\limits_{h\to0+}{f(h)\over h} = \lim\limits_{h\to0-}{f(h)\over h}$

but ${f(h)\over h} = \sin{1\over h}$ does not converge and keep moving between $-1$ and $1$. Thus it is not locally linear.


How's my reasoning? If true, leave some elaboration, else please denote the logical error.

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    $\begingroup$ You have to define $g(x)=f(x)$ when $x\neq 0$, and $g(0)=0$. Then $g$ is continuous. $f$ cannot be continuous at $x=0$ because it is not defined there. Your argument showing that $g$ is not differentiable at $x=0$ is correct. $\endgroup$ – uniquesolution Jun 26 '17 at 16:43
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    $\begingroup$ Your reasoning is great (+1) But I don't like the term "local linear" at $x = 0$, it makes me think of some functions that are really linear in some neighborhood of $0$, and they have some nice property: math.stackexchange.com/questions/116964/… $\endgroup$ – Li Chun Min Jun 26 '17 at 17:07
  • $\begingroup$ @LiChunMin your critics actually have some similarity with mine. One thing I feel curious is that.. my book denotes "local linear at x" is a equal term with "differentiable at x". But I've never seen this "local linear" description nowhere except this book $\endgroup$ – Beverlie Jun 26 '17 at 17:10
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To prove that $$\lim_0\sin (\frac 1h) $$ doesn't exists, it is equivalent to show that $$\lim_{+\infty}\sin (x) $$ doesn't exist since it is an odd function.

for this, observe that

$$\lim_{n\to+\infty}\sin (n\pi )=0$$ and $$\lim_{n\to+\infty}\sin (\frac \pi 2+2n\pi)=1.$$

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  • $\begingroup$ However your proof only applicable in case of sinx, however how could one show that xsin1/x is "not" defined at x =0? One should show that the value uniquely exist to show that well-definiteness. However I am still confused how to show the "existence" in this case. $\endgroup$ – Beverlie Jun 28 '17 at 7:09
  • $\begingroup$ it is between -x and x so it goes to zero by squeeze theorem. $\endgroup$ – hamam_Abdallah Jun 28 '17 at 9:55

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