
#1
May2409, 06:15 AM

P: 41

1. The problem statement, all variables and given/known data
does lim x →0 sin(1/x) exist? 3. The attempt at a solution I am very new to calculus. From what i have understood till now, limit of a function at a point exists only if the left and right hand limits are equal. What would the graph of sin(1/x) look like? Would the left and right hand limits exist? Do they exist in all cases, for all functions? 



#2
May2409, 06:44 AM

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P: 9,398

Alternatively we can think about lim sin(y) as y tends to infinity. 



#3
May2409, 06:49 AM

P: 197

Btw, the existence of left and right limits of a function at a certain point is necessary but not sufficient for a limit to exist at that point. Left and right limits should also be equal for a limit to exist at the same point.




#4
May2409, 07:07 AM

Math
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PF Gold
P: 38,904

Does lim x →0 sin 1/x exist? also.....
In particular, think about the sequences [itex]x_n= 1/(n\pi)[/itex], [itex]x_n= 2/((4n+1)\pi)[/itex], and [itex]x_n= 2/((4n1)\pi)[/itex] a n goes to infinity.




#5
May2409, 08:28 AM

P: 116

In a real analysis class you would do something like the following:
Let [tex] f(x) = sin ( \frac {1}{x} ) [/tex] Let [tex] s_{n} [/tex] be the sequence [tex] \frac {2}{n \pi} [/tex] [tex] \lim_{n \to \infty}s_{n} = 0 [/tex] but [tex] (f(s_{n}) )[/tex] is the sequence 1,0,1,0,1,0,1,0... which obviously doesn't converge. Therefore, [tex] \lim_{x \to 0} sin( \frac {1}{x}) [/tex] does not exist. 



#6
May2409, 10:23 AM

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P: 26,167

Hi hermy!
matt's is the best way for general questions like this … it shows you what the difficulty is! 



#7
May2509, 01:30 AM

P: 41





#8
May2509, 01:59 AM

Mentor
P: 14,483

f(x) = 0 if x is rational, 1 if it is irrational.




#9
May2509, 02:09 AM

Mentor
P: 21,081

Another example is f(x) = 1/x, x [itex]\neq [/itex] 0; f(0) = 0.
This function is defined for all real numbers. The left hand limit, as x approaches zero is negative infinity, while the right hand limit, as x approaches zero, is positive infinity. 



#10
May2509, 04:14 AM

P: 4,570

If we consider a left hand limit if sin x/x then we can verify the limit with a table of calculations (just grab a calculator and calculate f(x) sin x / x for values approaching x) We can also consider the right hand limit also. For the right hand limit we can do the same thing by letting f(x) approach sin x/x. Now the limit is only valid if and only if the right hand limit equals the left hand limit. So essentially if this is the case and you find a value for it, then you can prove what that limit is. 



#11
May2509, 04:38 AM

Mentor
P: 14,483

The original poster is looking at f(x)=sin(1/x), not g(x)=sin(x)/x.




#12
May2509, 05:54 AM

Sci Advisor
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P: 9,398

f(x)=sin(1/x) x=/=0 f(0)=0 (or anything else you want) 


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