What's the difference between x approaches 0 and equals zero? I understand that 'x approaches zero' is a very significant term as it allows division by x. But at times it seems as if we are dividing by x using the fact that it isn't zero and then say it's zero to get the answer.
For instance,
(From Kline's Calculus: An intuitive and physical approach)
$k = 128h + 16 h^2$  
Now, dividing both sides by h. (Which is allowed as h is never taken to be $0$.)
$k/h = 128 + 16h$
Now as h approaches $0$ the term $16h$ becomes vanishingly small. Thus:
$k/h = 128$  
I'm very sorry if I'm just missing an important point in understanding this concept. But...it still feels like I claimed it's only approaching 0 and not equal to 0, solved the issue of division and then said it's 0, but I don't have any division any more so problem solved. I certainly am confused with the concept of approaching a value and equal to a value. Any help on the concept of limits would be much appreciated. Thank you very much.
So if we simply ignore the vanishing term, is calculus like an approximation?
 A: So you've stumbled upon the concept of a limit and how it can be different than evaluating an expression at the value you are approaching.
When we are asking, for example, what the limit of $2 \cdot\frac{x-1}{x-1}$ is as $x$ approaches $1$, what we aren't asking for is what the value of the expression $2 \cdot\frac{x-1}{x-1}$ is at $x = 1$.  This expression is undefined at $x = 1$ since you get $\frac{0}{0}$.
But what we want to know is: as you choose $x$ closer and closer to the value $1$, are the values of $2 \cdot\frac{x-1}{x-1}$ getting closer and closer to some value?
Well, as it turns out in my example, it's easy to see that, yes, the values are getting closer to some value.  If $x$ is not equal to $1$, $2 \cdot\frac{x-1}{x-1}$ is equal to $2$.  So as $x$ gets closer and closer $1$, the expression $\frac{x-1}{x-1}$ is "getting closer and closer" to $2$, since it's already always $2$ for all $x$, which implies it's always $2$ for all $x$ near $1$ (except of course at $x=1$).
So a limit doesn't care about what happens at the value (e.g., at $x=1$).  It cares about what the expression looks like as $x$ gets closer and closer to $1$.
Sometimes, evaluating an expression agrees with its limit.  For example, consider the expression $x/2$.  What happens as $x$ gets closer and closer to $0$?  Well, if $x$ is getting really small, so is $x/2$.  So when $x$ is near $0$, $x/2$ is near $0$.  So we say $\lim \limits_{x \to 0} x/2 = 0$.  But if you evaluate $x/2$ at $x=0$, then we also get $0$.  When this happens, we say the expression is continuous at $x = 0$.  So $f(x) = x/2$ is continuous at $x = 0$.
In the case of your example, $(k/h) = 128 + 16h$ when $h$ is not zero.  So, since the $(k/h)$ equals $128 + 16h$ for all non-zero $h$, then as $h$ gets smaller, $(k/h)$ behaves as $128 + 16h$.  But $128 + 16h$ is getting closer and closer to $128$ as $h$ goes to $0$ since $16h$ is getting smaller and smaller.  That means $k/h$ is getting closer and closer to $128$.  So we write $\lim \limits_{h \to 0} k/h = 128$.  Note that this doesn't mean $k/0 = 128$.  As I discussed above, we don't care about what happens at the value of $h = 0$.  We only care about what happens as $h$ gets closer and closer to $0$.  $k/0$ is undefined.  But $\lim \limits_{h \to 0} k/h = 128$.
A: You use the fact that x is "approaching 0" for example to compute how to extend $f(x) = \frac{x^2}{x}$ by continuity to $0$. When $x$ is close to 0 but not 0, $\frac{x^2}{x}$ is a continuous function and therefore you can cross out two $x$, yielding $x$. Therefore the expected extension would be to make $f(0) = 0$ since when $x$ is really really small, $f(x) = x$.
So just take $x_k = 10^{-k}$ with $k \rightarrow \infty$. $x$ is never 0 so $f(x_k) = x_k$ but as $x_k$ gets really really close to 0, so does $f(x_k)$.
Hence the utility of considering $x$ as approaching $a$: when assuming $x=a$ would give some problem, try to show that when $x$ is close to $a$ some property holds "decently". Then expect it to hold as well when $x=a$ and apply it as well.
A: If $|h| $ is "very,very small", then we have that $k/h $ is very near by $128$, but we have not that $k/h = 128$
A: you have
$$\lim_{h\to0}k=\lim_{h\to0}(128h+16h^2)=0.$$
then
$$\lim_{h\to0}\frac{k}{h}=\frac{0}{0}$$
which is an indeterminate form.
but, if we simplify by $h$, we get
$$\lim_{h\to0}\frac{k}{h}=\lim_{h\to0}(128+16h)=128.$$
A: When you conider the above equation let h tends to zero,where h is 0.0000----1 .When you square this you get a number 0.00000000--------1,which is many times small than 
h.So we neglect higher power terms which does not mean that h^n is 0 (n>1).The final result is approximately equal but not the exact value
