Exercise: Let $(a_n)_{n=m}^\infty$ be a sequence of real numbers. Let $L$ and $L'$ be distinct real numbers. If $\lim_{n\to \infty}a_n=L$ and $\lim_{n\to \infty}a_n=L'$. Prove that $L=L'$.
Proof: Given that $\lim_{n\to \infty}a_n=L$ and $\lim_{n\to \infty}a_n=L'$, for every $\varepsilon > 0$ there exists an $N\ge m$ such that for all $n>N$ $$|a_n-L|<\frac \varepsilon 2$$ $$|a_n-L'|=|L'-a_n|<\frac \varepsilon 2$$
Adding the two inequalities together and using the triangle inequality we get $$|a_n-L+L'-a_n|=|-L+L'|<\varepsilon$$
Because $\varepsilon$ was an arbitrary real number $>0$, the above inequality implies that $|-L+L'|=|0|$ and thus $L=L'$.
Is this proof correct?
Edit: I thought I should type this proof better for any future readers through the suggestions that I got.
Proof: Given that $\lim_{n\to \infty}a_n=L$ and $\lim_{n\to \infty}a_n=L'$, for every $\varepsilon > 0$ there exists an $N_1\ge m$ and an $N_2 \ge m$ such that for all $n>max\{N_1,N_2\}$ we have the following two inequalities. $$|a_n – L|<\frac \varepsilon 2$$ $$|a_n – L'|=|L'-a_n|<\frac \varepsilon 2$$
Adding the two inequalities together and using the triangle inequality we get $$|a_n-L+L'-a_n|=|-L+L'|<\varepsilon$$
Because the above inequality must hold for every $\varepsilon > 0$, $|-L+L'|$ cannot be non-zero. Thus $|-L+L'|=|0|$ and $L=L'$.
Best Answer
It's not necessarily the same $N$ for $L$ and for $L'$. Otherwise it's all good.