Evaluate the limit. $$ \lim _{x \rightarrow 1} \frac{\ln x}{20 x-x^{2}-19}=\square $$ IType an exact answer.)

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To evaluate the limit
$$
\lim_{x \rightarrow 1} \frac{\ln x}{20x - x^2 - 19},
$$
we can follow these steps:

1. **Check the Indeterminate Form:**
   - As $$ x \rightarrow 1 $$:
     - $$ \ln(1) = 0 $$.
     - $$ 20(1) - 1^2 - 19 = 0 $$.
   - So, the expression is of the indeterminate form $$ \frac{0}{0} $$, allowing us to use L'Hôpital's Rule.

2. **Apply L'Hôpital's Rule:**
   - Differentiate the numerator and the denominator with respect to $$ x $$:
     - Numerator derivative: $$ \frac{d}{dx} (\ln x) = \frac{1}{x} $$.
     - Denominator derivative: $$ \frac{d}{dx} (20x - x^2 - 19) = 20 - 2x $$.
   
   - The limit becomes:
     $$
     \lim_{x \rightarrow 1} \frac{\frac{1}{x}}{20 - 2x}.
     $$

3. **Evaluate the New Limit:**
   - Substitute $$ x = 1 $$ into the expression:
     $$
     \frac{\frac{1}{1}}{20 - 2(1)} = \frac{1}{18}.
     $$

**Final Answer:**
$$
\frac{1}{18}
$$
The limit is $$ \frac{1}{18} $$.

Evaluate the limit.
IType an exact answer.)

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Evaluate the limit. IType an exact answer.)