Proving that $\lim_{n\to \infty}a_n=L$ and $\lim_{n\to \infty}a_n=L'$ imply $L=L'$. 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 $L=L'$.
 A: It's not necessarily the same $N$ for $L$ and for $L'$. Otherwise it's all good.
A: I think your solution is fine. Well done!
But, to be more precise, I would proceed as follows.
Since $(a_{n})_{n\in\mathbb{N}}$ is supposed to converge to both $L$ and $L'$, for every $\varepsilon > 0$, there corresponds $n_{1}(\varepsilon)$ and $n_{2}(\varepsilon)$ such that $|a_{n} - L| < \varepsilon$ for every $n\geq n_{1}(\varepsilon)$ and $|a_{n} - L'| < \varepsilon$ for every $n\geq n_{2}(\varepsilon)$.
Consequently, if $L\neq L'$ and $\varepsilon = |L - L'|/3$, one can choose $n(\varepsilon) = \max\{n_{1}(\varepsilon),n_{2}(\varepsilon)\}$ so that for every $n\geq n(\varepsilon)$ one gets the following relation:
\begin{align*}
|L' - L| = |(a_{n} - L) - (a_{n} - L')| \leq |a_{n} - L| + |a_{n} - L'| \leq 2\varepsilon \Rightarrow |L - L'| \leq 0 \Rightarrow L = L'
\end{align*}
which contradicts the original assumption.
Observation
Although you have chosen the same $N$ for both limits, there is no harm at doing so because it is the same as taking the maximum between $n_{1}(\varepsilon)$ and $n_{2}(\varepsilon)$ as I have proposed in my solution.
Hopefully this helps!
