Suppose f: D ---->R with $\displaystyle x_0$ an accumulation point of D. Assume L1 and L2 are limits of f at $\displaystyle x_0$. Prove L1=L2.

My thought is to assume L1 is not equal to L2, but I'm stuck here.

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- Nov 8th 2008, 01:58 PMkathrynmathL1 and L2 limits of f. Prove L1=L2
Suppose f: D ---->R with $\displaystyle x_0$ an accumulation point of D. Assume L1 and L2 are limits of f at $\displaystyle x_0$. Prove L1=L2.

My thought is to assume L1 is not equal to L2, but I'm stuck here. - Nov 8th 2008, 03:30 PMJhevon
Ok, since they are not equal, we have $\displaystyle |L_1 - L_2| > 0$

Now choose $\displaystyle \epsilon = \frac {|L_1 - L_2|}3$

Then $\displaystyle |L_1 - L_2| = |L_1 - L_2 + f(x) - f(x)| = \dots$

now continue and try to get a contradiction on the magnitude of $\displaystyle |L_1 - L_2|$ - Nov 8th 2008, 03:34 PMThePerfectHacker
Here is an outline of proof you need to fill in more details to make it formal.

$\displaystyle |L_1 - L_2| = |(f(x) - L_1) - (f(x) - L_2)| \leq |f(x) - L_1| + |f(x) - L_2| < \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon$

Since this is for any $\displaystyle \epsilon > 0$ it forces $\displaystyle |L_1 - L_2| = 0 \implies L_1 = L_2$.