## is 0 to the power of infinity indeterminate

It is indeterminate, and the value depends on how you are getting the $\infty$ and the $0$. There is no universal value for $\infty^0$. But notice that we know the denominator of the exponent will not, however small, ever by zero, so that the exponent is of the form 0/b with b non-zero. It says "infinity to the zeroth power". if limx->a f(x) = 0 and limx->a g(x) = infinity, show that limx->a [f(x)]g(x) = 0 this shows that 0 to the infinite power is not an indeterminate form. Since the answer is ∞ 0, then it is also another type of Indeterminate Form and it is not accepted as a final answer in Mathematics.We know that any number raised to zero power is always equal to one except for infinity that's why it is also an Indeterminate Form. Both of these are called indeterminate … I might have missed a few. Some other indeterminate forms are $0^0, 1^\infty, \infty\times0,\frac00, 1$. Example 3: 0 0 0^0 0 0 represents the empty product (the number of sets of 0 elements that can be chosen from a set of 0 elements), which by definition is 1. This is also the same reason why anything else raised to the power of 0 is 1. Show us, how exactly do you proceed to carry out the multiplication, and what are the digits of your number 'infinity'? In particular, infinity is the same thing as "1 over 0", so "zero times infinity" is the same thing as "zero over zero", which is an indeterminate form. 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 \({\infty }/{-\infty }\;\) (recall that as \(x\) goes to infinity a polynomial will behave in the same fashion that its largest power behaves). It's basically the exponential equivalent of dividing by 0 (taking something to the 1/infinitieth power): Yes, this is a good equivalence. Your title says something else than "infinity times zero". suppose f is a positive function.

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