Prove the following statement: if $\lim _{n\to \infty }a_n = L $ and $\lim _{n\to \infty }a_n = M$, then $L=M$ I was going through some practice problems for an upcoming calculus exam and I came across the question: "Prove the following statement: if $\lim _{n\to \infty }a_n = L $ and $\lim _{n\to \infty }a_n = M$, then $L=M$". Here is what I tried to do:$$$$
Assume $L\neq M$. Therefore, $L=M+k$ where $k$ is some real value. So we can rewrite $$\lim _{n\to \infty }a_n - \lim _{n\to \infty }a_n = L-M = M+k-M =k$$ $$\lim _{n\to \infty }a_n - \lim _{n\to \infty }a_n =\lim _{n\to \infty }(a_n-a_n) = \lim _{n\to \infty }0 = 0 $$ $$\therefore k=0 $$ From there we can conclude from subbing $k=0$ that $L=M+0=M $ and this contradicts the statement $L\neq M$ so by contradiction $L=M$ must be true. $$$$
Is there anything wrong with this proof? I know for proving the uniqueness of limits theorem you have to do more but that isn't exactly what the question is asking.
 A: The proof is subtly circular, but there is something that can be salvaged from it. As it turns out, yes, you can actually apply algebra of limits to show that $0 = a_n - a_n \to L - M = k$. The proof of the algebra of limits theorem doesn't rely on uniqueness at all; that is if $a$ is one of the possibly many limits of $a_n$, and $b$ is one of the possibly many limits of $b_n$, then $a - b$ is one of the possibly many limits of $a_n - b_n$.
But, in your last conclusion, you say because $0$ is a limit of $0$ and $k$ is a limit of $0$, then $0 = k$. This assumes what you want to prove, but instead of assuming it for the arbitrary convergent sequence $a_n$, you are now assuming it about the very nice constant sequence $0$. While the proof is circular so far, it is what I would consider progress!
We just now need to show that the only limit of the $0$ sequence is $0$. I will let you think about how to do this. Why does the constant $0$ sequence not get any $\varepsilon$-close to any other $k \neq 0$? It would have to, say, get $|k|/2$-close. Why is this impossible?
A: I have thought immediately to the limit uniqueness theorem for sequences that asserts:

A $\{a_n\}$ sequence of real numbers cannot have two distinct limit $L$ and $M$. In fact if we suppose that $L\ne M$ are (finite; in truth, one can easily remove this restriction) limits of the $\{a_n\}$, we will show that $L=M$.

By the definition of a limit, for every $\varepsilon> 0$ (real number) exist $\nu_1$ and $\nu_2$ such that for every $n>\nu_1$ it is true $|a_n-L|<\varepsilon/2$, and for every $n> \nu_2$ it is true $|a_n-M|<\varepsilon/2 $ . Let $N=\max\{\nu_1, \nu_2\}$ then for every $n > N$
$$|L-M|\leq|L-a_n|+|a_n-M| <\varepsilon$$
for the  triangular inequality.  So $|L-M| <\varepsilon$ for every $\varepsilon >0$, and so $|L-M|=0$. Hence $$L=M, \qquad \blacksquare$$
A: As noticed your proof is not correct. Assuming by contradiction as you did $L\neq M$, we can proceed by the definition
$$\lim _{n\to \infty }a_n = L \iff \forall\varepsilon \quad \exists n_L \quad \forall n>n_L \quad |a_n-L|<\varepsilon$$
$$\lim _{n\to \infty }a_n = M \iff \forall\varepsilon \quad \exists n_M \quad \forall n>n_M \quad |a_n-M|<\varepsilon$$
and then fix $\varepsilon \le \frac{|M-L|}3$.
Can you conclude form here?
