# Extra Credit Problem from a past test

1. Nov 23, 2008

### shrug

Messed this up.

Last edited: Nov 24, 2008
2. Nov 23, 2008

### 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.

3. Nov 24, 2008

### Pere Callahan

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.

4. Nov 24, 2008

### HallsofIvy

Staff Emeritus
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. 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.