How can I prove this limit? \(\displaystyle\lim\lim{i}{t}{s}_{{{x}\to\infty}}\sqrt{{{x}^{{2}}-{4}{x}+{5}}}={x}-{2}\)

Step 1
Let me start with what you did wrong: This expression
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{x^2-4x+5}}$
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 ${\displaystyle \mathrm{\infty }}$). 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
${\displaystyle \mathrm{\infty }=x-2}$
It is only true if ${\displaystyle x=\mathrm{\infty }}$, which is not a value that normally x is allowed to even take. (That is why we talk about the "limit as ${\displaystyle x\to \mathrm{\infty }}$'' and not the "value at ${\displaystyle x=\mathrm{\infty }}$'') So your question as stated didn't make sense.
What should you have written? You actually meant that the two functions
${\displaystyle f\left(x\right)=\sqrt{x^2-4x+5}}$ and g(x) are asymptotic as ${\displaystyle x\to \mathrm{\infty }}$ There are different ways to define this. The comments have it interpreted as
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}[\sqrt{x^2-4x+5}-(x-2)]=0}$
This is somewhat tricky to show. herb steinberg almost has it, but doesn't justify the claim that it is ${\displaystyle x-2+O\left(\frac{1}{x}\right)}$
What you should start with is a change of variable: let ${\displaystyle t=x-2}$, noting that ${\displaystyle t\to \mathrm{\infty }}$ as ${\displaystyle x\to \mathrm{\infty }}$ and vice versa. Then it becomes
${\displaystyle \underset{t\to \mathrm{\infty }}{lim}\sqrt{t^2+1}-t}$
${\displaystyle =\underset{t\to \mathrm{\infty }}{lim}\frac{\sqrt{1+\frac{1}{t^2}}-1}{\frac{1}{t}}}$
${\displaystyle =\underset{s\to 0+}{lim}\frac{\sqrt{1+s^2}-1}{s}=f^{\prime }\left(0\right)}$
where ${\displaystyle f\left(s\right)=\sqrt{1+s^2}}$ It is easily checked that ${\displaystyle f^{\prime }\left(0\right)=0}$
A somewhat looser definition is
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\left[\frac{\sqrt{x^2-4x+5}}{x-2}\right]=1}$
Which is easy to prove, since for ${\displaystyle x>2}$
${\displaystyle \frac{\sqrt{x^2-4x+5}}{x-2}=\sqrt{1+\frac{1}{{(x-2)}^2}}}$