Does lim x →0 sin 1/x exist? also..... 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?
Re: Does lim x →0 sin 1/x exist? also..... try drawing it and find out. It would be a bit dull if left and right limits always exist. Remember that limits are unique. So, you need to think about sequences of real numbers x_n tending to zero. Alternatively we can think about lim sin(y) as y tends to infinity.
Re: Does lim x →0 sin 1/x exist? also..... 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.
Re: 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/((4n-1)\pi)[/itex] a n goes to infinity.
Re: Does lim x →0 sin 1/x exist? also..... 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.
Hi hermy! matt's is the best way for general questions like this … it shows you what the difficulty is!
Re: Does lim x →0 sin 1/x exist? also..... Could you give an example of a function where the left/right hand limit doesn't exist at a point (where the function is defined on the set of all real no.s)?
Re: Does lim x →0 sin 1/x exist? also..... 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.
Re: Does lim x →0 sin 1/x exist? also..... In calculus you need to use the idea of limits. Essentially the limit of sin x/x does equal 1 but you have to show it from both sides. 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.
Re: Does lim x →0 sin 1/x exist? also..... The original poster is looking at f(x)=sin(1/x), not g(x)=sin(x)/x.
Re: Does lim x →0 sin 1/x exist? also..... I don't want to give the game away but.... sin(1/x) as x tends to 0 perhaps? I can trivially extend this to a function on the whole of the real line if I want to, if you're worried about it being undefined at zero: f(x)=sin(1/x) x=/=0 f(0)=0 (or anything else you want)