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

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 $(-\infty, 0) \cup (5, \infty)$.

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:

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.

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 $\text{Does not exist}$.