Why can you cancel zero terms in the numerator and denominator of limits? [closed]

There's a factor of $(x-2)$ in the denominator, and two factors of $(x-2)$ in the numerator. The cancelling removes one factor of each. What remains is

$$\frac{x-2}{x+2}$$

As $x$ approaches $2$, the numerator approaches $0$ and the denominator approaches $4$. This is a well defined quotient: $0/4=0$, which gives you the answer.

The key thing to note here is that limits consider what happens near $x=2$, but not at $x=2$. Near $x=2$, $(x-2)$ is not $0$ and hence the cancellation is valid.

This is also a good example that limits involving division by zero need not diverge to infinity. In fact, they can diverge to $-\infty$ or $+\infty$, converge to anything in between, or even have some other, more wild behavior.

Big Picture.

1) If $f(x) = \frac {(x-2)^2}{(x-2)(x+2)}$ then $f(0) = \frac 00$ which is undefined (not $\infty$ by the way; simply undefined and meaningless).

However and values very very close to $x =2$ but not actually precisely $x=2$ all the $f(x)$ will have values that are very close to a value $k$. That is what we mean when you say $\lim\limits_{x\to 2}f(x) = k$. We are saying "if $x$ is close to but not necessarily equal to$2$, then $f(x)$ is close to but not necessarily equal to $k$

Of course, we have to be technically precise and formal about it (which the above most certainly is not).

so.... we are looking at values of $x$ where $x$ is NOT equal to $2$.

2). If $M = \frac {AB}{AC}$ and $A \ne 0$ then we can divide $M = \frac {\not AB}{\not A C} = \frac AC$.

And if $x$ is not equal to $2$ then $x -2 \ne 0$. So we can "cancel" $x-2$ out.

So for $x \ne 2$ then $f(x) =\frac{(x-2)^2}{(x-2)(x+2)} \frac {\not{(x-2)}(x-2)}{\not{(x-2)}(x+2)} = \frac {x-2}{x+2}$.

This new expression at $x = 2$ will be $\frac {2-2}{2+2} = \frac 04 = 0$.

This is not $f(2)$ (which is still undefined) but an all $x$ very close but not equal to $2$ then $f(x) = \frac {(x-2)(x-2)}{(x-2)(x+2)}$ is very close to but not equal to $0$.