I'm sorry to ask this question , really there are many indeterminates cases in
the computation of limit of numerical functions such that :$\infty-\infty $, $0^0$ , $0*\infty$ $\cdots $ .
My question here is:
Is $\sqrt{-\infty}$ one of the indeterminate cases ?
Note: Really I would like to know more about the mathematical notion of indeterminate cases !!!!!
Thank you for any kind of help
On
This is not indeterminate. Note that for any positive real number $x$, $\sqrt{-x}$ is imaginary. Thus if you have some expression that is approaching $-\infty$ under a square root, then essentially, you are approaching infinity along the imaginary axis of the complex plane. As such, the magnitude of $\sqrt{-\infty}$ is always $\infty$.
The indeterminate forms are ones where the limit can have different values depending on how the problematic numbers are reached - for example, a $0/0$ limit could be something like $x/x$ where the numerator and denominator approach 0 at the same rate, so they effectively cancel out; or it could be more like $x^2/x$ where the numerator approaches 0 "faster" and the limit goes to zero, or it could be more like $x/x^2$ where the denominator goes to 0 faster, so the limit has no bound (i.e. it's "infinite").
By comparison, if you have a limit where you're taking the square root of a function that goes negative, then you just have undefined values. Since there's no sensible value to give as you approach the limit point (assuming we're working in the real numbers), the limit does not exist.