Limits and algebraic simplification 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
$$= \lim_{x \to \infty} x(\sqrt {1 - 0} - 1)=\lim_{x \to \infty} x(1 - 1)=0 $$

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.
