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April 2nd, 2018, 09:06 AM  #1 
Member Joined: Jan 2016 From: Blackpool Posts: 97 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 09:11 AM. 
April 2nd, 2018, 09:15 AM  #2  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 226 Thanks: 76 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
Fix $\epsilon > 0$ and set $\delta = \epsilon$. Then whenever $0 < x0 < \delta$, we have $f(x)  0 = f(x) \leq x = x0 < \delta = \epsilon$. So $f$ is continuous at $0$.  
April 3rd, 2018, 04:47 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 nonzero 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, 05:00 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,403 Thanks: 2477 Math Focus: Mainly analysis and algebra  
April 10th, 2018, 01:24 PM  #5 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,462 Thanks: 106 
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 10th, 2018, 11:43 PM  #6  
Senior Member Joined: Oct 2009 Posts: 498 Thanks: 164  Quote:
Quote:
 
April 11th, 2018, 01:57 AM  #7 
Senior Member Joined: Aug 2012 Posts: 2,010 Thanks: 574  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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