## University Calculus: Early Transcendentals (3rd Edition)

(a) $\lim_{x\to2}f(x)=5$ (b) $\lim_{x\to2}f(x)=5$
(a) $$\lim_{x\to2}\frac{f(x)-5}{x-2}=3$$ Here we cannot use the substitution method like normal, since we use that method, the denominator would end up with $0$. Instead, we can rewrite $f(x)$ as follows: $$f(x)=(f(x)-5)+5$$ $$f(x)=\frac{f(x)-5}{x-2}\times(x-2)+5$$ Therefore, $$\lim_{x\to2}f(x)=\lim_{x\to2}\Big(\frac{f(x)-5}{x-2}\times(x-2)+5\Big)$$ $$\lim_{x\to2}f(x)=\lim_{x\to2}\frac{f(x)-5}{x-2}\times\lim_{x\to2}(x-2)+\lim_{x\to2}(5)$$ $$\lim_{x\to2}f(x)=3\times(2-2)+5$$ $$\lim_{x\to2}f(x)=3\times0+5=5$$ (b) $$\lim_{x\to2}\frac{f(x)-5}{x-2}=4$$ Again, we rewrite $f(x)$ into: $$f(x)=\frac{f(x)-5}{x-2}\times(x-2)+5$$ Therefore, $$\lim_{x\to2}f(x)=\lim_{x\to2}\frac{f(x)-5}{x-2}\times\lim_{x\to2}(x-2)+\lim_{x\to2}(5)$$ $$\lim_{x\to2}f(x)=4\times(2-2)+5$$ $$\lim_{x\to2}f(x)=4\times0+5=5$$