# Show that a limit of a sequence is unique

Ok I have the proof in front of me, there is just one step I don't understand and im hoping someone here can clarify it.

Basically it asks to show that each convergent series has only one limit. So you start off by supposing $a_n$ is convergent, and $L_1$ and $L_2$ are limits.

Let $\epsilon\gt 0$. Then there exists some $N$ such that $|a_n - L_1| \lt \frac{\epsilon}{2}$ and $|a_n - L_2| \lt \frac{\epsilon}{2}$ for all $n>N$.

Then $|L_1-L_2| \le |a_n - L_1| + |a_n - L_2 | < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$. Since $\epsilon \gt 0$ is arbitrary, $|L_1-L_2| = 0$, so $L_1 = L_2$. I get the last step, the step I am confused about is

$|L_1-L_2| \le |a_n - L_1| + |a_n-L_2|$. Can anyone provide clarity here?

• Triangle inequality. – Pedro Tamaroff Apr 10 '14 at 1:40

This is a standard trick of adding $0$ creatively and using the triangle inequality:
$$|L_1 - L_2| = |L_1 - a_n + a_n - L_2| = \left|\Big(L_1 - a_n\Big) + \Big(a_n - L_2\Big)\right| \le |L_1 - a_n| + |a_n - L_2|$$