Please help me out. Is there exist such a number like 0^(k), where k<0

Question

What is the value of $0^x$ where 'x' is negative real number and what will be the value of $(x)^{1/x}$ as $x$ tends to zero.

Answer 1

Let $0^{-k}$ for some $k > 0$.

If you take $0^{-k}$ to be the multiplicative inverse of $0$ raised to the power of $k$, that is $0^{-k}= (0^{-1})^k$, such a number will not be defined because there is no multiplicative inverse of $0$.

If you take $0^{-k}$ to be the multiplicative inverse of $0^k$, that is $0^{-k} = (0^k)^{-1}$, such a number will also not be defined because $0^k = 0$ and similarly, there is no multiplicative inverse of $0$.

Simply put, $0^x$ does not make sense for a $x \in \mathbb{R}^{-}$.

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Also, the limit as $x \to 0$ for $x^\frac{1}{x}$ does not exists because the left and right limits are not equal. In fact, the left limit doesn't even exists in $\mathbb{R}$.