Extra Credit Problem from a past test

[shrug]

Messed this up.

 

[Dick]

I don't know. I think you'd better post the full question. I can't make much sense out of that. I can't even figure out what the actual question is.

 

[Pere Callahan]

Prove or give a counterexample.
Xn and Yn are two diferent sequences that --> to c. If F(Xn) and F(Yn) both converge to L, then Lim (Fx) =L as X-->c. Didn't get any thing at all. How do you approach this. Or does it exist at all. If the Limit exists then wouldn't it equal L. Thanks

If the limit exists it must sure be L. Just define F by F(Xn)=F(Yn)=L and F(x)=0 else. Then F certainly does not have a limit as x-->c.

 

[HallsofIvy]

I think what you are asking is this: if \lim_{n\rightarrow\infty}x_n= c and [/itex]\lim_{y_n\rightarrow 0} y_n= c[/itex], \lim_{n\rightarrow\infty}f(x_n)= L, and \lim_{n\rightarrow\infty}f(y_n)= L, is it necessarily true that \lim_{x\rightarrow c}f(x)= L?

The answer is no. What is true is that \lim_{x\rightarrow c} f(x)= L if and only if \lim_{n\rightarrow\infty} f(x_n}= L for every sequence {x_n} that converges to c. Just two sequences isn't enough.

For example, suppose f(x)= 0 if x is rational, 1 if x is irrational. Let {x_n} and {y_n} be two different sequences of rational numbers converging to 1. Then \lim_{n\rightarrow\infty} f(x_n)= \lim_{n\rightarrow\infty} g(x_n)= 0 but \lim_{x\rightarrow 1} f(x) does not exist.<br /> <br /> Of course, if we know that\lim_{x\rightarrow c} f(x)[/itex] exists then one such sequence is sufficient to tell us what the limit is.