Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 55

Assume $\displaystyle \lim \limits_{x \to 1} \frac{f(x) - 8}{x- 1} = 10$, Find $\lim \limits_{x \to 1} f(x)$


$
\begin{equation}
\begin{aligned}
& \lim \limits_{x \to 1} \frac{f(x) - 8}{x - 1} = 10
\qquad
\Longrightarrow
\qquad
\frac{\lim \limits_{x \to 1} [f(x) - 8]}{\lim \limits_{x \to 1}(x-1)} = 10
\qquad
\text{(Multiplying $\lim \limits_{x \to 1} (x - 1)$ to both sides of the equation)}\\


& \frac{\lim \limits_{x \to 1} [f(x) - 8]}{\cancel{\lim \limits_{x \to 1}(x-1)}}
\cdot
\cancel{\lim \limits_{x \to 1}(x-1)}
= 10 \cdot \lim \limits_{x \to 1}(x-1)\\

& \lim \limits_{x \to 1} [f(x) - 8] = 10 \cdot \lim \limits_{x \to 1}(x-1)\\

& \lim \limits_{x \to 1}f(x) - \lim \limits_{x \to 1} 8 = 10 \cdot \lim \limits_{x \to 1} (x-1)\\

& \lim \limits_{x \to 1} f(x) - 8 = 10 \cdot (1-1)\\

& \lim \limits_{x \to 1} f(x) = 0 +8 = 8\\

& \lim \limits_{x \to 1} f(x) = 8

\end{aligned}
\end{equation}
$