Answer to Question #260167 in Calculus for Alunsina

Question

Find lim h→0 f(h+3)-f(3)/h

a. f(x) = x²-3

b. f(x) = x/x-2

c. f(x) = sqrt x+1

Answer 1

a. f(x)=x²-3

f(h+3)=(h+3)²-3=h²+6h+9-3=h²+6h+6

f(3)=3²-3=9-3=6

$\lim\limits_{h\rarr 0}\frac{f(h+3)-f(3)}{h}=\lim\limits_{h\rarr 0}\frac{(h^2+6h+6)-(6)}{h}$

$=\lim\limits_{h\rarr 0}\frac{h^2+6h+6-6}{h}$

$=\lim\limits_{h\rarr 0}\frac{h^2+6h}{h}$

$=\lim\limits_{h\rarr 0}\frac{h(h+6)}{h}$

$=\lim\limits_{h\rarr 0}{h+6}$

=0+6

=6

b. f(x)=$\frac{x}{x-2}$

f(h+3)=$\frac{h+3}{h+3-2}=\frac{h+3}{h+1}$

f(3)=$\frac{3}{3-2}=\frac{3}{1}=3$

$\lim\limits_{h\rarr 0}\frac{f(h+3)-f(3)}{h}=\lim\limits_{h\rarr 0}{\frac{\frac{h+3}{h+1}-3}{h}}$

$\lim\limits_{h\rarr 0}{\frac{\frac{h+3}{h+1}-3}{h}}=\lim\limits_{h\rarr 0}{\frac{\frac{h+3-3(h+1)}{h+1}}{h}}$

=$\lim\limits_{h\rarr 0}{\frac{\frac{h+3-3h-3}{h+1}}{h}}$

=$\lim\limits_{h\rarr 0}{\frac{\frac{h-3h}{h+1}}{h}}=\lim\limits_{h\rarr 0}({{\frac{h-3h}{h+1}}*\frac{1}{h}})$

=$\lim\limits_{h\rarr 0}{{\frac{h-3h}{(h+1)h}}}=\lim\limits_{h\rarr 0}{{\frac{h(1-3)}{h(h+1)}}}$

$=\lim\limits_{h\rarr 0}{{\frac{1-3}{h+1}}}$

$=\frac{-2}{0+1}=\frac{-2}{1}$

=-2

c. f(x)=$\sqrt{x+1}$

f(h+3)=$\sqrt{h+3+1}=\sqrt{h+4}$

f(3)=$\sqrt{3+1}=\sqrt{4}=2$

$\lim\limits_{h\rarr 0}\frac{f(h+3)-f(3)}{h}=\lim\limits_{h\rarr 0}\frac{\sqrt{h+4}-2}{h}$

$\lim\limits_{h\rarr 0}\frac{\sqrt{h+4}-2}{h}=\lim\limits_{h\rarr 0}(\frac{\sqrt{h+4}-2}{h}*\frac{\sqrt{h+4}+2}{\sqrt{h+4}+2})$

$=\lim\limits_{h\rarr 0}\frac{h+4+2\sqrt{h+4}-2\sqrt{h+4}-4}{h(\sqrt{h+4}+2)}$

$=\lim\limits_{h\rarr 0}\frac{h}{h(\sqrt{h+4}+2)}=\lim\limits_{h\rarr 0}\frac{1}{\sqrt{h+4}+2}$

$=\frac{1}{\sqrt{0+4}+2}=\frac{1}{\sqrt{4}+2}$

$=\frac{1}{2+2}$

=$\frac{1}{4}$