How can I prove this limit?
$$\lim \limits_{x \to \infty} \sqrt{x^2-4x+5} = x-2$$
Graphically, it is easy to see that this is true, but is it possible to prove it mathematically?
EDIT:
I'm sorry, it appears that I worded my question wrong? I don't quite understand what I should do to correct it, but I will try to rephrase instead.
I'm trying to mathematically prove that for big values of $x$ the function $f(x)=\sqrt{x^2-4x+5}$ starts to resemble the linear function $g(x)=x-2$. Is this possible?
And as a side note, I would love to thoroughly know what I did wrong the first time, I've read all the comments, but I still don't quite understand. Thank you.
Let me start with what you did wrong: This expression $$\lim_{x \to \infty} \sqrt{x^2-4x+5}$$ does not depend on a variable $x$. There is no $x$ in it. The $x$ you see is a dummy variable. That $x$ exists only within that notation, and is there only to make the notation work. In computing, they call that "scope". $x$ comes into existence with the "lim" and goes back out of existence after the $5$ and the closing of the square root.
The expression represents a single value (though in this case, not a member of the real numbers, but $\infty$). It is not a variable expression. It is not a function. It is a single value. On the other hand, "$x - 2$" is a variable expression. It takes on different values for different values of $x$, which is a free variable here. It has no relation to the dummy variable $x$ that appears on the left and does not exist on the right.
So what you wrote reduces to $$\infty = x - 2$$ It is only true if $x = \infty$, which is not a value that normally $x$ is allowed to even take. (That is why we talk about the "limit as $x\to \infty$" and not the "value at $x = \infty$".) So your question as stated didn't make sense.
What should you have written? You actually meant that the two functions $f(x) = \sqrt{x^2-4x+5}$ and $g(x)$ are asymptotic as $x \to \infty$. There are different ways to define this. The comments have it interpreted as $$\lim_{x \to \infty} \left[\sqrt{x^2-4x+5} - (x-2)\right] = 0$$
This is somewhat tricky to show. herb steinberg almost has it, but doesn't justify the claim that it is $x - 2 + O(1/x)$. What you should start with is a change of variable: let $t = x - 2$, noting that $t \to \infty$ as $x \to \infty$ and vice versa> Then it becomes $$\lim_{t \to \infty} \sqrt{t^2+1} - t = \lim_{t \to \infty} \frac{\sqrt{1+\frac1{t^2}} - 1}{\frac1t}= \lim_{s \to 0+} \frac{\sqrt{1+s^2} - 1}s = f'(0)$$ where $f(s) = \sqrt{1+s^2}$. It is easily checked that $f'(0) = 0$.
A somewhat looser definition is $$\lim_{x \to \infty} \left[\frac{\sqrt{x^2-4x+5}}{x-2}\right] = 1$$ Which is easy to prove, since for $x > 2$, $$\frac{\sqrt{x^2-4x+5}}{x-2} = \sqrt{ 1 + \frac 1{(x-2)^2}}$$