To evaluate the limit
[tex]$
\lim_{x \rightarrow 1} \frac{\ln(x)}{11x - x^2 - 10},
$[/tex]
we can start by checking if this limit is in an indeterminate form of the type [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex].
First, substitute [tex]\(x = 1\)[/tex] into the numerator and the denominator:
- The numerator [tex]\(\ln(1) = 0\)[/tex].
- The denominator [tex]\(11 \cdot 1 - 1^2 - 10 = 11 - 1 - 10 = 0\)[/tex].
Since both the numerator and the denominator are zero, we have an indeterminate form [tex]\(\frac{0}{0}\)[/tex].
In this case, l'Hôpital's Rule is applicable, which states that for limits of the form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], we can take the derivatives of the numerator and the denominator:
[tex]$
\lim_{x \rightarrow 1} \frac{\ln(x)}{11x - x^2 - 10} = \lim_{x \rightarrow 1} \frac{\frac{d}{dx}[\ln(x)]}{\frac{d}{dx}[11x - x^2 - 10]}.
$[/tex]
Now, calculate the derivatives of the numerator and the denominator:
- The derivative of the numerator [tex]\(\ln(x)\)[/tex] is [tex]\(\frac{1}{x}\)[/tex].
- The derivative of the denominator [tex]\(11x - x^2 - 10\)[/tex] is [tex]\(11 - 2x\)[/tex].
Thus, applying l'Hôpital's Rule, we get:
[tex]$
\lim_{x \rightarrow 1} \frac{\ln(x)}{11x - x^2 - 10} = \lim_{x \rightarrow 1} \frac{\frac{1}{x}}{11 - 2x}.
$[/tex]
Now, simplify the expression inside the limit:
[tex]$
\lim_{x \rightarrow 1} \frac{\frac{1}{x}}{11 - 2x} = \lim_{x \rightarrow 1} \frac{1}{x(11 - 2x)}.
$[/tex]
Substitute [tex]\(x = 1\)[/tex] into this simplified expression:
[tex]$
\frac{1}{1 \cdot (11 - 2 \cdot 1)} = \frac{1}{11 - 2} = \frac{1}{9}.
$[/tex]
Therefore, the limit is:
[tex]$
\boxed{\frac{1}{9}}.
$[/tex]
