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?
Best Answer
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 $\lim \limits_{h \to 0} k/h = 128$.