The periodic functions f(x): R -> R and g(x): R -> R satisfy the condition:
lim (f(x)-g(x))=0 for all x ε R
Prove that f(x)=g(x) for every real number x
try to prove this. trust me it is really a good one.
must try
62I think bhargav means this
The periodic functions f(x): R -> R and g(x): R -> R satisfy the condition:
lim (f(x)-g(x))=0 for all a ε R
x->a
Prove that f(x)=g(x) for every real number x
Bhargav can you confirm this?
62bhargav.. not every one in the world is as smart as u :P
for students in class XII i am sure that not more than 1 in 1000 know tihs definition.
39The proof uses the definiton that for every sub-sequence of the sequence, the limit tends to zero
isnt it obvious?
62In the proof of bhargav, the problem for most ppl will be that the definition of limits is not very obvious.
The proof uses the definiton that for every sub-sequence of the sequence, the limit tends to zero.... that is what he has used.
39here`s a solution:
Suppose T1 and T2 are the periods of f and g respectively. For a fixed x lim(f(x+nT1)-g(x+nT1)) =0 as n->infinity, so lim g(x+nT1)=f(x) as n->infinity. By the same way lim f(x+nT2)=g(x) as n->infinity. We also have lim(f(x+nT1+nT2)-g(x+nT1+nT2)) =0 as n-> infinity, so lim(f(x+nT2)- g(x+nT1))=0 as n->infinity, but this limit is g(x)-f(x) (this is for a fixed x), so g(x)-f(x)=0 for any x.
39sorry i slighly misunderstood the question myself.
KAYMANT SIR`S QUESTION IS RIGHT.
9superb counter example sir
learnt a lot from this thread abt continuity and limits
this is a classic
62look at the functions
f(x) = 0
g(x) = 0 if x is not an integer and 1 otherwise
lim x-> a f(X)-g(x) = 0 for all values of a.
Hence the question that I guessed is incorrect :P
I guess Anant sir's question is the correct one :)
9sir but it seems so obvious ( i may be wrong also)
lim h(x) =0 for all x
now if there isnt a continuity at x
wont the limit at x→x+ and x→x- not exist ??
9but here
bhargav has given that lim exists for each and every point
66well I have seen a very similar problem, it goes:
Show that if f,g : R → R are continuous and periodic and lim_{x→∞} (f(x)-g(x))=0, then f=g.
I think this is what Bhargav means.
66for a limit to exist at a point the function is not required to be continuous. in fact, it need not be defined at that point. consider as an example f(x)=(sin x)/x
What about the limit at x=0 where f is not even defined.
62lim (f(x)-g(x))=0 for all x ε R
this statement is a bit vague
Bhargav has not given where x limits to !
I think what he has written makes not much sense mathematically
Bhargav can you check if you have missed out somthing!?
66no celestine, your conclusion presupposes continuity of h(x) but that is not mentioned.
66@ b555, I think the limit you are talking about should be x→∞. Additionally, the continuity of these functions are required as well. (I am saying so because provided these I have a proof.)
