Limit as $x$ approaches 2 is undefined? Does following function have a limit if x approaches 2. Calculate what the limit is and motivate why if it is missing.
$$
\frac{(x-2)^2}{(x-2)^3} =\frac{ 1 }{ x-2}. 
$$
I answered $\frac{1 }{ 0 }= 0 $ undefined is that correct?
 A: It looks like you are considering the function
$$
f(x) = \frac{(x-2)^2}{(x-2)^3} = \frac{1}{x-2}.
$$
You want to consider what happens to this function when $x$ approaches $2$. Note that the numerator is just the constant $1$ and when $x$ approaches $2$, then $x - 2$ approaches $0$. So you have something that approaches $1$ divided by $0$. This limit does not exist (as you correctly state).
Note, however, that $1$ divided by $0$ is not equal to $0$. In fact $1$ divided by $0$ is undefined which is the reason that the limit is undefined.
If you consider the limit as $x$ approaches $0$ from the right, then you are just considering what happens to $1 / x$ for positive values of $x$. And since you are taking this (non-zero) constant and dividing it by something that becomes smaller and smaller (while being positive), then the limit is $\infty$:
$$
\lim_{x\to 0^+} f(x) = \infty.
$$
Likewise
$$
\lim_{x\to 0^-} f(x) = -\infty.
$$
A: The denominator approaches zero and the numerator doesn't, so the limit "does not exist". If you've seen the term "indeterminate", then the reason is that 1/0 is undefined rather than indeterminate.  However it's important to note that 1/0 is not 0, it's undefined.
