## Calculus 8th Edition

$\displaystyle \lim_{x\rightarrow 1}f(x)=\lim_{x\rightarrow 1}\{[f(x)-8]+8\}$ ... 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 $\displaystyle \lim_{x\rightarrow 1}f(x)=0+8=8$