Indeterminate forms, limits confusion

I'm rather new to limits and lately I been told about the indeterminate form which is sometimes attained when solving a limit by direct substitition. However, the inderterminate form has cause me some confusion. Could someone please answer this question;

1. Why is the indeterminate form special? Why do we try using factoring, conjugation or trig identies when we run into this form instead of giving up and concluding theres no limit. And if there is a limit how do we know. When solving a limit using direct substitution and arriving in a situation where the answer is a/0, where a is not zero we conclude the limit does not exist. Why do we not do this with indeterminate form, in both forms we have 0 as our denominator. I saw some online arguments saying that a/0 will result in a asymptote, but why does 0/0 not result in one?

P.S. I'm still new to Calculus so please do not use any complex terms or expressions. I'm also a self learner and have been following this course online. https://www.khanacademy.org/math/ap-calculus-ab/ab-limits-new.

Thanks to anyone who took the time to read this!!!

Solution 1:

Let us take, as an example, the one suggested by @MrGandalfSauron.

For small positive arc/angle $x$ we see that the relatives sizes of $\sin x, x$ and $\tan x$ satisfy the inequality

$$ \sin x < x < \tan x$$

Dividing by $\sin x$ we get

$$ 1 <\frac{x}{\sin x}<\frac{1}{\cos x}$$

Now, as $x\to0^+$ we see that both $x\to0$ and $\sin x\to 0$, whereas $\dfrac{1}{\cos x}\to1$.

So, even though $\lim_{x\to0^+}\dfrac{x}{\sin x}$ is a limit of type $\dfrac{0}{0}$, the limit is $1$, since it gets 'squeezed' between a value of $1$ and a value approaching $1$.

This is just one example. There exist other examples where limits of type $\dfrac{0}{0}$ approach other values, or fail to exist altogether. This is why such limits are said to be indeterminate. Each must be taken on a case-by-case basis.

Solution 2:

If $a_n\to \infty$ and $b_n\to\infty$ then we can conclude that $a_n+b_n\to\infty$. Therefore this is not an indefinite form; we always know what the result is. Similarly, if $a_n\to 0$ and $b_n\to 0$ then we can conclude for sure that $a_n+b_n\to 0$.

But if $a_n\to\infty$ and $b_n\to\infty$ then you can't conclude anything at all about $\frac{a_n}{b_n}$, because there is not enough data: it depends on $a_n$ and $b_n$. For instance, if $a_n = n^2$ and $b_n=n$ then $\frac{a_n}{b_n}=n\to\infty$, while if $a_n=n$ and $b_n=n$ then $\frac{a_n}{b_n}=1$, and so on. All this means is that if all I told you is that $a_n\to\infty$ and $b_n\to\infty$ then you can't conclude anything about $\frac{a_n}{b_n}$; but if I actually tell you who $a_n$ and $b_n$ are, then you can calculate the limit.

In the case of $a_n+b_n$ where $a_n\to\infty,~b_n\to\infty$, we could conclude that $a_n+b_n\to\infty$ without even knowing who $a_n$ and $b_n$ are. That is what makes this a determinate form, whereas the case of so called $\frac{\infty}{\infty}$ is indeterminate, in the sense that in order to calculate the limit we need more information on $a_n$ and $b_n$. Hopefully this makes intuitive sense: if I tell you that $a_n$ and $b_n$ are huge numbers then you know for sure that $a_n+b_n$ is also huge. But if I tell you that $a_n$ and $b_n$ are huge numbers then you don't know for sure what's going on with $\frac{a_n}{b_n}$; it depends which of $a_n$ and $b_n$ is bigger than the other.

In the same way, if $a_n\to 0$ and is positive, then for every positive real number $x$ we have that $\frac{x}{a_n}\to\infty$. Intuitively, this is because if you take some fixed number and divide it by some very tiny number, the result will be huge. However, if $a_n\to 0$ and $b_n\to 0$ then you don't know in general what's going on with $\frac{a_n}{b_n}$; the result will depend on $a_n$ and $b_n$ and in each and every case you will need to come up with some trick for calculating the limit. This is intuitively so because if $a_n$ and $b_n$ are very little then we don't know anything about $\frac{a_n}{b_n}$, it depends on which of them is smaller than the other.