Dealing with different limits on different sides of an equality This is related to another problem that I have, but I want to work on that one more without direct hints.
Consider the following:
$\lim_{n\to\infty} \left( 2 - \frac{1}{n} \right) = \lim_{x\to 2} x$
This is "clear" when we know the value of the limits, but what if we wanted to prove this assertion rigorously without prior knowledge of the limits in question? I feel as if there is some $\epsilon - \delta$ magic that has to take place, something along the lines of first proving that one of the limits exists, and then proving that:
$\left| f(x) - \lim_{n \to \infty} a_n \right|  < \epsilon $
But this feels extremely messy, and I have no idea how to deal with a limit in a normal expression. Alternatively I can just assign some value to the limit such as $c$ and then just work with that, but the problem is that in the problem I have I need to use the properties of the limiting function in question. Any ideas?
 A: The easiest way is probably to formally prove that both limits exist and have value $2$.
The proofs aren't very hard:
Let $\epsilon>0$, we choose $N=\frac{1}{\lfloor\epsilon\rfloor}$. Then for all $n>N$ we have $|2-\frac{1}{n}-2|<|\frac{1}{N}|=|\lfloor\epsilon\rfloor|\leq \epsilon$. 
So $\lim_{n\to\infty} (2-\frac{1}{n})=2$
Now let $\epsilon>0$, we choose $\delta=\epsilon$. Then if $|x-2|<\delta$, we have $|x-2|< \epsilon$.
So $\lim_{x\to 2} x=2$
We conclude that
$\lim_{n\to\infty} (2-\frac{1}{n})=2=\lim_{x\to 2} x$
A: While @Uncountable proved this particular equality, the asker clarified in a comment that the actual question was

I was hoping that an exact proof which isn't based around proving they both have the same value could shed some light on the actual problem I have. The actual problem I have has two different incalculable limits. So I cannot simply state that they are equal, but I need to prove via some properties that the two limits are equal.

One way to do that is to take approaching sequences for both limits and look at the differences of the values:
to show $\lim_{n → ∞} a_n = \lim_{n → ∞} b_n$ look at $|a_n - b_n|$ (or $d(a_n, b_n)$ for general metric complete spaces). If $\lim_{n → ∞} |a_n - b_n| = 0$, then you know that those limits are equal if they exist. For existence you need that your space (in your case $ℝ$, so OK) is complete and that $a_n$ and $b_n$ are Cauchy sequences, that is
$$
\lim_{k →∞} \sup_{n, m > k} |a_n - a_m| = 0
$$
$\lim_{x → 0} f(x)$ fits in this framework by
$$
\lim_{x → 0} f(x) = a ⇔ ∀(x_n)_n \text{ with } x_n → 0: \lim_{n→∞} f(x_n) = a
$$
I think you can look for the topics Cauchy sequence and completion for further reading.
