Vtyjkyuk

I really cant underatand what is exactly difference. By their it can be verified that undefined has no value at all but inderminant form we can able to evaluate its limit by various way(such as L'Hospital's rule) whether it exists.

Indeterminate forms generally refer to combining limits. If you have two functions $f(x)$ and $g(x)$ such that $lim_x to infty f(x)=a, lim_x to infty g(x)=b$ you can say $lim_x to infty f(x)+g(x)=a+b$. Other operations can be done similarly. It is also true that if $lim_x to infty f(x)=+infty, lim_x to infty g(x)=b$ you can say $lim_x to infty f(x)+g(x)=+infty$ if you define infinite limits in the usual way. Indeterminate forms come about when you have something like $lim_x to infty f(x)=+infty, lim_x to infty g(x)=-infty$ where you would like to say $lim_x to infty f(x)+g(x)=+infty-infty$. This is indeterminate because you can get any value. The simple combination is not sufficient and you need to look how fast $f$ and $g$ go off to $infty$ to come up with the result.

Undefined means exactly that. Division by zero is undefined. $0^0$ is undefined. This is not in reference to limits at all.

I tend to think of an indeterminate form as being one whose value is ambiguous in some way while leaving something undefined is done to avoid absurdity or trivialities. As for $0^0$ it's considered indeterminate because we have both $x^0=1$ and $0^x=0$ for $xneq 0$ and there is no clear means by which we can resolve this in a consistent manner.

For comparison we leave division by zero undefined because if we don't then taking $a=b=1$ we would have $a^2-b^2=a-b$ and so $(a+b)(a-b)=a-b$ leaving us with $a+b=2=1$. We can either conclude every number is zero or that division by zero leads to contradiction so it's left undefined.

Undefined is something that is not and will never be defined. e.g dividing by zero. Is just an expression that has no sense and no possible value. On the other side, indeterminate is an expression that you can't know its value at simple sight, but it can be anything (even undefined). Some indeterminate forms are $0/0$ or $infty/infty$. These kind of expressions are usually solvable using limits, and they can have any value.

In summary, undefined expressions are hopeless, indeterminate ones may have an answer if you work them.

The case $0^0$ is just another example of indeterminate form, and it can be worked in various ways usually taking natural log will bring the exponent down. For example:

$$lim_xto0 x^x=e^lim_xto0ln(x^x)=e^lim_xto0xln(x)=e^lim_xto0fracln(x)1/x$$

So now you can use LH:

$$e^lim_xto0frac1/x-1/x^2 = e^0=1$$