Answer
8
Work Step by Step
$\displaystyle \lim_{x\rightarrow 1}f(x)=\lim_{x\rightarrow 1}\{[f(x)-8]+8\}$
... Law 1, The limit of a sum...
$=\displaystyle \lim_{x\rightarrow 1}=[f(x)-8]+\lim_{x\rightarrow 1}8$
... Law 7. $\displaystyle \lim_{x\rightarrow a}c=c$
$=\displaystyle \lim_{x\rightarrow 1}[f(x)-8]+8\qquad(*)$
$f(x)-8=(f(x)-8)\displaystyle \cdot\frac{x-1}{x-1},\quad (x\neq 1)$
so
$\displaystyle \lim_{x\rightarrow 1}[f(x)-8]=\lim_{x\rightarrow 1}[\frac{f(x)-8}{x-1}\cdot(x-1)]$
... Law 4, The limit of a product...
$=\displaystyle \lim_{x\rightarrow 1}\frac{f(x)-8}{x-1}\cdot\lim_{x\rightarrow 1}(x-1)$
$=10\cdot 0$
$=0$
Inserting this into (*), we have
$\displaystyle \lim_{x\rightarrow 1}f(x)=0+8=8$
