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

To evaluate the limit \( \lim _{x \rightarrow 1} \frac{\ln x}{15 x - x^2 - 14} \), we first plug in \( x = 1 \): \[ \ln 1 = 0 \] \[ 15(1) - (1)^2 - 14 = 15 - 1 - 14 = 0 \] Since both the numerator and denominator approach 0, we have a \( \frac{0}{0} \) indeterminate form, which means we can apply l'Hôpital's Rule. Applying l'Hôpital's Rule, we differentiate the numerator and the denominator: 1. The derivative of \( \ln x \) is \( \frac{1}{x} \). 2. The derivative of \( 15x - x^2 - 14 \) is \( 15 - 2x \). Now, applying l'Hôpital's Rule, we now have: \[ \lim_{x \to 1} \frac{\ln x}{15 x - x^2 - 14} = \lim_{x \to 1} \frac{\frac{1}{x}}{15 - 2x} \] Now we evaluate this new limit by plugging in \( x = 1 \): \[ \frac{\frac{1}{1}}{15 - 2(1)} = \frac{1}{15 - 2} = \frac{1}{13} \] Thus, the final result is: \[ \lim _{x \rightarrow 1} \frac{\ln x}{15 x - x^2 - 14} = \frac{1}{13} \]