# Indeterminate forms

1. Jun 1, 2015

Here's a list of all the indeterminate forms I'm familiar with:
$\frac{0}{0}$, $\frac{\infty}{\infty}$, $0⋅\infty$, $\infty - \infty$, $0^0$, $1^{\infty}$, $\infty^0$
Suppose we want to evaluate the limit:
$$\lim_{x→0} x^2 \cos{\frac{1}{x}}$$
We can find the value of this limit by applying the squeeze theorem. The limit would otherwise be indeterminate; if we plug in $x = 0$, we get:
$$0⋅\cos{\frac{1}{0}}$$
Under what category does this indeterminate form lie?

2. Jun 1, 2015

### ShayanJ

This is not an indeterminate form. It has a unique limit: 0.
The reason is, whatever the argument of the cosine is, we know that the result lies between -1 and 1. So we know that we are multiplying zero by a number between -1 and 1 which is obviously zero.

3. Jun 2, 2015

I'm confused. The limit of, say, $x^3 - x^2$ as $x$ approaches infinity is also unique; it's infinity. However, at first glance, it seems to be indeterminate ($\infty - \infty$).
I only arrived at the result by factoring out $x^3$.
Perhaps the definition of "indeterminate" is not so clear to me.

4. Jun 2, 2015

### ShayanJ

The difference is that, in the case of indeterminate forms, at first you encounter one of the seven forms. Then you should find a way to see what's the value of the indeterminate form in this particular example. Its sometimes finite and sometimes infinite and sometimes doesn't exist. About your example, we first encounter $\infty - \infty$ which is indeterminate i.e. we don't know its value. Then we search for an alternative way to find its value and we figure out that because $x^3$ rises faster than $x^2$, at infinity they should be very far apart which means the answer is infinity.
But about $\displaystyle \lim_{x\to 0} x^2 \cos{\frac 1 x}$, we can directly find out that the answer is zero. Because cosine is always between -1 and 1, and here is no different. We just don't know what is the value of the cosine but we know its between -1 and 1. So just call its value a. Then we have $\displaystyle \lim_{x\to 0} x^2 a$ which is obviously zero.

5. Jun 2, 2015

### ShayanJ

This particular function has limit, as x goes to infinity of 0, but there are other functions, such as $x^2- (x^2+ 1)$, that are "of the form" $\infty- \infty$ that converge to other limits. "$\infty- \infty$" and "0/0" are "indeterminate" because you cannot determine such a limit by just setting $x= 0$ or $x= \infty$ in the sequence.