# L1 and L2 limits of f. Prove L1=L2

• Nov 8th 2008, 02:58 PM
kathrynmath
L1 and L2 limits of f. Prove L1=L2
Suppose f: D ---->R with $x_0$ an accumulation point of D. Assume L1 and L2 are limits of f at $x_0$. Prove L1=L2.

My thought is to assume L1 is not equal to L2, but I'm stuck here.
• Nov 8th 2008, 04:30 PM
Jhevon
Quote:

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

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

Ok, since they are not equal, we have $|L_1 - L_2| > 0$

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

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

now continue and try to get a contradiction on the magnitude of $|L_1 - L_2|$
• Nov 8th 2008, 04:34 PM
ThePerfectHacker
Quote:

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

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

Here is an outline of proof you need to fill in more details to make it formal.
$|L_1 - L_2| = |(f(x) - L_1) - (f(x) - L_2)| \leq |f(x) - L_1| + |f(x) - L_2| < \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon$

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