To solve the limit [tex]\(\operatorname{Lim}_{x \rightarrow 1} \frac{x^2 + 81}{x - 9}\)[/tex], follow these steps:
1. Substitute [tex]\( x = 1 \)[/tex]:
Plugging [tex]\( x = 1 \)[/tex] directly into the function, we get:
[tex]\[
\frac{1^2 + 81}{1 - 9} = \frac{1 + 81}{-8} = \frac{82}{-8} = -\frac{82}{8} = -\frac{41}{4}
\][/tex]
2. Conclusion:
Thus, the limit of the function as [tex]\( x \)[/tex] approaches 1 is:
[tex]\[
\boxed{-\frac{41}{4}}
\][/tex]
This concludes that [tex]\(\operatorname{Lim}_{x \rightarrow 1} \frac{x^2 + 81}{x - 9} = -\frac{41}{4}\)[/tex].
