In the first limit if we plugged in x=4 we would get 0/0 and in the second limit if we “plugged” in infinity we would get ∞/−∞ (recall that as x goes to infinity a polynomial will behave in the same fashion that its largest power behaves). Is Infinity to the power of infinity indeterminate? An indeterminate form is a limit that is still easy to solve. It's like how 0*0 is not indeterminate form. If you raise a number of magnitude less than one to a higher power, the result is smaller. Before proceeding with examples let me address the spelling of “L’Hospital”. It only means that in its current form as a limit put into a function, it presents too many unknowable characteristics to form an appropriate answer properly. So, L’Hospital’s Rule tells us that if we have an indeterminate form 0/0 or ∞/∞ ∞ / ∞ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit. Both of these are called indeterminate forms.. The more modern spelling is … If you raise a smaller number to the same power, the result is smaller. The functions resulting in 0/0 and infinity over negative infinity can achieve a solution through various means. Thus you'll get a limit of zero no matter which order you take it in, and the result is completely unambiguous.
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