Dilemma about the value of $\frac{4- 4}{4 - 4}$ I can't find where the mistake is here. Can someone explain how it is possible?


The expression $\frac{4-4}{4-4}$ has no well-defined value because it is just $\frac{0}{0}$.
I like the trick of factorising the numerator and denominator to give $\frac{(2-2)(2+2)}{2(2-2)}$. 
However, when you try to cancel $(2-2)$ from both the numerator and denominator, you must ask yourself what it means to "cancel". When you "cancel the twos" to give $\frac{2}{6} = \frac{1}{3}$, what have you done? You have divided both the numerator and denominator by 2.
When you try to cancel the $(2-2)$ in your fraction, you are dividing both the numerator and denominator by $(2-2)$. The problem is $2-2=0$, and so you have divided both the numerator and denominator by zero. It's bad enough dividing by zero once, but you did it twice!
This kind of cancellation can work when you are dealing with limits. For example, what happens to $\frac{x^2}{x}$ as $x$ gets very small. It makes sense for all $ x \neq 0$. For example, when $x=3$, $\frac{x^2}{x}=\frac{9}{3}=3$. When $x=2$, $\frac{x^2}{x}=\frac{4}{2}=2$. When $x=1$, $\frac{x^2}{x}=\frac{1}{1}=1$. What about $x=0$? When $x=0$, $\frac{x^2}{x}$ is not well-defined and takes no value. However, we can first pretend that $x \neq 0$ and get $\frac{x^2}{x} \equiv x$. Then as $x \to 0$, we say that 
$$\lim_{x \to 0} \frac{x^2}{x} = \lim_{x \to 0} x = 0$$
