My Math Forum Continuity of this function

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 April 2nd, 2018, 10:06 AM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 100 Thanks: 2 Continuity of this function f(x)=xsin(1/x) for f(x)$\neq 0$ and f(0)=0 I am struggling to understand why my teacher claims that the function is continuous at f(0). My teacher says since |f(x)|$\leq$|x| for all x and is therefore continuous however i do not understand what this is trying to say, is it a sandwhich/squeeze theorem proof? Please could someone help me with my intuition, Thanks edit: also why is this function not differentiable at x=0, if we take the difference quotient we get f'(0)=$\frac{xsin(1/x)}{x}=sin(1/x)$ where sin(1/x) is undefined at x=0, is this correct? Last edited by Jaket1; April 2nd, 2018 at 10:11 AM.
April 2nd, 2018, 10:15 AM   #2
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Quote:
 Originally Posted by Jaket1 My teacher says since |f(x)|$\leq$|x| for all x and is therefore continuous however i do not understand what this is trying to say, is it a sandwhich/squeeze theorem proof?
This lets you do an easy epsilon-delta proof.

Fix $\epsilon > 0$ and set $\delta = \epsilon$. Then whenever $0 < |x-0| < \delta$, we have $|f(x) - 0| = |f(x)| \leq |x| = |x-0| < \delta = \epsilon$. So $f$ is continuous at $0$.

 April 3rd, 2018, 05:47 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 Yes, what your teacher is saying is that you can use a 'squeeze' method. sin(1/x) is always between -1 and 1 for any non-zero x so $\displaystyle -1\le sin(1/x)\le 1$. If $\displaystyle x> 0$ then $\displaystyle -x\le x sin(1/x)\le x$ and taking the limit as x goes to 0 from above, we "squeeze" the limit to 0. If $\displaystyle x< 0$ then $\displaystyle -x\ge x sin(1/x)\ge x$ so that again the limit is 0.
April 3rd, 2018, 06:00 AM   #4
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 Originally Posted by Jaket1 why is this function not differentiable at x=0, if we take the difference quotient we get f'(0)=$\frac{xsin(1/x)}{x}=sin(1/x)$ where sin(1/x) is undefined at x=0, is this correct?
Yes.

 April 10th, 2018, 02:24 PM #5 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 Lim f(x) exists but f(0) doesn't unless you define f(0) to be lim f(x) which makes f(x) continuous at x=0. I think that's the real point of OP, not the limit. Derivative doesn't have to exist at a point because it is continuous there. Sharp point for ex. As for why it doesn't exist here, f(x) is constantly oscilating with period approaching zero so you can't nail down the slope.
April 11th, 2018, 12:43 AM   #6
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 Originally Posted by zylo Lim f(x) exists but f(0) doesn't unless you define f(0) to be lim f(x) which makes f(x) continuous at x=0. I think that's the real point of OP, not the limit.
The OP defined f(0) = 0.

Quote:
 Derivative doesn't have to exist at a point because it is continuous there. Sharp point for ex. As for why it doesn't exist here, f(x) is constantly oscilating with period approaching zero so you can't nail down the slope.
While it is true that the derivative of f at 0 does not exist, it is not because of the reason you gave. Take $f(x) = x^2 \sin(1/x)$ for $x\neq 0$ and $f(0) = 0$. This function is differentiable, but $f$ is also constantly oscillating with period approaching zero.

April 11th, 2018, 02:57 AM   #7
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 Originally Posted by Jaket1 f(x)=xsin(1/x) for f(x)$\neq 0$ and f(0)=0 Please could someone help me with my intuition
The intuition is that $\sin$ is bounded no matter how much it oscillates as $x$ gets close to 0. It stays between -1 and 1. And $x$ goes to zero, so $x \sin(\frac{1}{x})$ goes to zero.

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