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Step-by-Step Solution
Step 1: Understand the Expression
We want to evaluate the limit
$ \displaystyle \lim_{x \to 0} \frac{\log\bigl(x^n\bigr) - [x]}{[x]}, $
where $n \in \mathbb{N}$ and $[x]$ denotes the greatest integer less than or equal to $x$ (also known as the floor function). The expression involves the behavior of the floor function as $x$ approaches $0$.
Step 2: Consider the Behavior of $[x]$ as $x \to 0^+$
When $x$ approaches $0$ from the positive side (i.e., $x \to 0^+$ with $x > 0$ but very small), we have $0 < x < 1$, so $[x] = 0$. Hence, in the expression
$ \frac{\log\bigl(x^n\bigr) - [x]}{[x]}, $
the denominator becomes $[x] = 0$, which already suggests a problematic scenario. Moreover, $\log(x^n) = n \log(x)$ goes to $-\infty$ as $x \to 0^+$, but since $[x] = 0$, we must be careful. However, the denominator being $0$ itself signals issues with the limit from the right side.
Step 3: Consider the Behavior of $[x]$ as $x \to 0^-$
When $x$ approaches $0$ from the negative side (i.e., $x \to 0^-$ with $x < 0$ and very close to $0$), we have $-1 < x < 0$, so $[x] = -1$. Therefore, near $0$ from the left, the denominator in
$ \frac{\log\bigl(x^n\bigr) - [x]}{[x]} $
is $[x] = -1$ (not $0$). But $\log(x^n)$ is undefined for negative $x$ if we stay within the realm of real numbers (unless $n$ is odd, even then it is tricky for a limit approaching $0$ through negative values). Clearly, the left-hand side behavior and the right-hand side behavior differ drastically for the floor function.
Step 4: Conclude the Non-existence of the Limit
The value of $[x]$ “jumps” from $-1$ to $0$ as we cross $0$ from the left to the right. This discontinuity in the denominator means the overall expression cannot have a unique limit. Even before considering deeper subtleties, the denominator $[x]$ does not approach a single finite value as $x$ approaches $0$. Hence, the limit
$ \lim_{x \to 0} \frac{\log(x^n) - [x]}{[x]} $
does not exist.
Final Answer
The limit does not exist.
