Please let me know what is wrong in the following algebraic simplification which gives a wrong limit value.
$$\lim_{x \to \infty} (\sqrt {x^2 - 4x} - x)=\lim_{x \to \infty} (\sqrt {x^2 (1 - 4/x)} - x) $$
$$= \lim_{x \to \infty} (x\sqrt {1 - 4/x} - x)= \lim_{x \to \infty} x(\sqrt {1 - 4/x} - 1) $$
When $x$ tends to infinity, $4/x$ will be negligible and hence
$$= \lim_{x \to \infty} x(\sqrt {1 - 0} - 1)=\lim_{x \to \infty} x(1 - 1)=0 $$
This is a very common mistake and frankly speaking I don't understand the reason why students tend to make such mistakes. Perhaps most people believe that there are no rules in calculus. Like in algebraic manipulation one is aware of the fact that addition is commutative but division is not, there are rules for limits also and one has to work according to those rules only. There is no specific rule which allows us to replace $4/x$ with $0$. So such manipulation is not justified. The idea that limits can be evaluated using hand waving techniques based on arguments like "something is negligible compared to something else" has to be ditched seriously.
Remember that the meaning of the equation $$\lim_{x\to\infty} \frac{4}{x}=0$$ is not that whenever you see expression $4/x$ (as part of a limit evaluation as $x\to \infty$) you can replace it by $0$ but rather the equation simply means that whenever you see the expression $\lim\limits_{x\to\infty} \dfrac{4}{x}$ you can replace it by $0$.
Also note that the usual technique for this problem is to multiply by conjugate to get the limit as $$-\lim_{x\to\infty} \frac{4}{\sqrt{1-(4/x)}+1}$$ Contrary to what many beginners believe we don't just ignore $4/x$ and get the answer as $-2$. Rather we work according to laws of algebra of limits. First step is to analyze the denominator. Note that we have $$\lim_{x\to\infty} 1-\frac{4}{x}=1- \lim_{x\to\infty} \frac{4}{x}=1-0=1$$ and since the square root function is continuous at $1$ we get $$\lim_{x\to\infty} \sqrt{1-(4/x)}=\sqrt{\lim_{x\to\infty} (1-(4/x))}=\sqrt{1}=1$$ And then $$\lim_{x\to\infty}\sqrt{1-(4/x)}+1=1+1=2$$ Thus the denominator tends to a non-zero limit and hence the desired limit is $-4/2$. All of these steps are justified by a well known rule of algebra of limits. The entire working looks like we have just ignored $4/x$ but in reality a lot of laws are working behind the scenes to give the effect of ignoring $4/x$.
If we don't multiply by conjugate and instead focus on the expression $$\lim_{x\to\infty} x(\sqrt{1-(4/x)}-1)$$ then we have a problem as we can't use the product rule of limits to write the above as $$\lim_{x\to\infty} x\cdot\lim_{x\to\infty} (\sqrt{1-(4/x)}-1)$$ and we have no further way to proceed unless we apply some algebraic manipulation on the expression under limit.
Normally one does not need to bother with all these details while evaluating limits in step by step manner. But we do need to keep in mind the limitations of algebra of limits. You may have a look at this answer which gives the crux of these limit laws in a very useful/simple form.
Note that $$\lim_{x\to a} f(x)g(x) = \lim_{x \to a} f(x) \cdot \lim_{x\to a} g(x)$$ when both limits ($\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$) exist and are finite. In your case, we cannot apply the above formula as one of the limits is infinity.
$$\lim_{x \to \infty} x(\sqrt {1 - 4/x} - 1) \neq \lim_{x \to \infty} x \cdot \lim_{x \to \infty}(\sqrt {1 - 4/x} - 1).$$
The issue is that while $\sqrt{1-\frac4x}-1$ is getting close to $0$, the factor $x$ is getting unboundedly large. So it is unclear what will happen. This type of limit is called an indeterminate form for just that reason.
A simpler example of the same sort of thing is:
$\lim_{x\to\infty}(x\cdot \frac1x)$.
The second factor, $\frac1x$, goes to $0$ (just as in your example), but the first factor gets unboundedly large. Clearly in this example, the limit is $1$, not $0$.
In your attempt you are reaching to a relation of the form $0 \times \infty$ which is not even defined.
So you cannot conclude that is is zero.
But you can do this if you want:
$\sqrt{x^2-4x}-x=\frac{(\sqrt{x^2-4x}-x)(\sqrt{x^2-4x}+x)}{\sqrt{x^2-4x}+x}=\frac{-4x}{\sqrt{x^2-4x}+x}=\frac{-4}{\sqrt{1-\frac{4}{x}}+1} \to -2$
as $x \to +\infty$
By the Taylor development, $\sqrt{1-\epsilon}\approx1-\dfrac\epsilon2$. Hence keeping this $\epsilon$, the limit is $-2$.
Declaring a term negligible was premature.
Errors aside, you could also consider $x=1/h$, which gives
$$L=\lim_{h\to0}\frac{\sqrt{1-4h}-\sqrt1}h=\frac d{dt}\sqrt{4t}\bigg|_{t=1}=2\frac d{dt}\sqrt t\bigg|_{t=1}$$
By graphing, it is clear this limit exists and is non-zero, as the derivative is clearly non-zero. By applying the power rule, one can also see this limit is equal to 1.
TL;DR: "When $x$ tends to infinity, $4/x$ will be negligible and hence [...]" is not justified algebraic simplification.
We could just as well say: "When $x$ tends to infinity $\sqrt{1-\frac 4x}-1$ is negligible..." That said, just replace $\sqrt{1-\frac 4x}-1$ with any $f(x)$ such that $\lim_{x\to\infty} f(x) = 0$.
With the above adjustments, we could state your reasoning as:
$$\lim_{x\to\infty}f(x) = 0\implies \lim_{x\to\infty} xf(x) = 0.$$
It doesn't take much creativity to come up with counterexample, take $f(x) = \frac 1x$.
This mistake is similar to another common mistake: $\lim_n(1+\frac 1n)^n = \lim_n 1^n = 1$.
You cannot simply ignore the fact that when taking limit of $f(x)g(x)$ that $f$ and $g$ change simultaneously and not independent of one another. The rule that $$\lim_{x\to a} (f(x)g(x)) = \lim_{x\to a} f(x)\cdot \lim_{x\to a} g(x)$$ when the limits on RHS exist is a piece of magic that we take for granted.
