Evaluate the following limit. Use l'Hôpital's Rule when it is convenient and applicable. $$ \lim _{x \rightarrow 1} \frac{\ln x}{15 x-x^{2}-14} $$

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To evaluate the limit
$$
\lim_{x \rightarrow 1} \frac{\ln x}{15x - x^{2} - 14},
$$
we can use **L'Hôpital's Rule** because the limit initially presents an indeterminate form of $$\frac{0}{0}$$.

1. **Check the Indeterminate Form:**
   - As $$x \rightarrow 1$$:
     - $$\ln(1) = 0$$
     - $$15(1) - (1)^2 - 14 = 15 - 1 - 14 = 0$$
   - This gives the form $$\frac{0}{0}$$, which allows us to apply L'Hôpital's Rule.

2. **Differentiate the Numerator and Denominator:**
   - Numerator derivative: $$\frac{d}{dx} \ln x = \frac{1}{x}$$
   - Denominator derivative: $$\frac{d}{dx} (15x - x^{2} - 14) = 15 - 2x$$

3. **Apply L'Hôpital's Rule:**
   - The limit becomes:
     $$
     \lim_{x \rightarrow 1} \frac{\frac{1}{x}}{15 - 2x}
     $$
   - Substitute $$x = 1$$ into the simplified expression:
     $$
     \frac{\frac{1}{1}}{15 - 2(1)} = \frac{1}{13}
     $$

Therefore, the limit is:

$$
\boxed{\dfrac{1}{13}}
$$
The limit is $$\frac{1}{13}$$.

Evaluate the following limit. Use l'Hôpital's Rule when it is convenient and applicable. \( \lim _{x \rightarrow 1} \frac{\ln x}{15 x-x^{2}-14} \)

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