lim x->0 ln(1/(x2-5x))

To evaluate the limit, we first need to analyze the domain of the function and the behavior of its argument as $x$ approaches $0$.

Step 1: Determine the domain of the function. The natural logarithm $\ln(u)$ is defined only when $u > 0$. In this case, $u = \frac{1}{x^2 - 5x}$. So, we must have $\frac{1}{x^2 - 5x} > 0$. This implies that $x^2 - 5x$ must be positive. Factoring the denominator, we get $x(x-5) > 0$. This inequality holds when both factors are positive ($x > 0$ and $x-5 > 0 \implies x > 5$) or when both factors are negative ($x < 0$ and $x-5 < 0 \implies x < 0$). Thus, the domain of the function is $(-\infty, 0) \cup (5, \infty)$.

Step 2: Evaluate the left-hand limit as $x \to 0^-$. As $x$ approaches $0$ from the left side ($x \to 0^-$), $x$ is a small negative number. For example, if $x = -0.001$, then $x-5 = -5.001$. So, $x(x-5)$ is (negative) $\times$ (negative) = positive. Therefore, as $x \to 0^-$, $x^2 - 5x \to 0^+$. This means $\frac{1}{x^2 - 5x} \to \frac{1}{0^+} = +\infty$. Then, the left-hand limit is: $\lim_{x \to 0^-} \ln\left(\frac{1}{x^2 - 5x}\right) = \lim_{y \to \infty} \ln(y) = \infty$

Step 3: Evaluate the right-hand limit as $x \to 0^+$. As $x$ approaches $0$ from the right side ($x \to 0^+$), $x$ is a small positive number. For example, if $x = 0.001$, then $x-5 = -4.999$. So, $x(x-5)$ is (positive) $\times$ (negative) = negative. Therefore, as $x \to 0^+$, $x^2 - 5x \to 0^-$. This means $\frac{1}{x^2 - 5x} \to \frac{1}{0^-} = -\infty$. Since the natural logarithm is not defined for negative arguments, $\ln(-\infty)$ is undefined. Thus, the right-hand limit $\lim_{x \to 0^+} \ln\left(\frac{1}{x^2 - 5x}\right)$ does not exist.

Step 4: Conclude the overall limit. For the overall limit $\lim_{x \to 0} f(x)$ to exist, both the left-hand limit and the right-hand limit must exist and be equal. Since the right-hand limit does not exist (the function is undefined for values of $x$ approaching $0$ from the right), the overall limit does not exist.

The final answer is $\boxed{Does not exist}$.

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