Question

${lim}_{h\to 0}[\frac{1}{h3\sqrt{8+h}}-\frac{1}{2h}]=$

• A. $\frac{-1}{12}$
  
• B. $\frac{-4}{3}$
• C. $\frac{-16}{3}$
  
• D. $\frac{-1}{48}$
  

Answer 1

The correct option is D

$\frac{-1}{48}$

${lim}_{h\to 0}[\frac{1}{h3\sqrt{8+h}}-\frac{1}{2h}]=$
$={lim}_{h\to 0}\left[\frac{1}{h}\{\frac{1}{3\sqrt{8+h}}-\frac{1}{2}\}\right]$
$={lim}_{h\to 0}\left[\frac{1}{h}\left\{\frac{2-(8+h{)}^{\frac{1}{3}}}{2\times 3\sqrt{8+h}}\right\}\right]$
$={lim}_{h\to 0}\left[\frac{1}{h}\left\{\frac{8^{\frac{1}{2}-(8+h{)}^{\frac{1}{3}}}}{2\times 3\sqrt{8+h}}\right\}\right]$
$[A^2-B^2=(A-B\left)\right(A^2+AB+B^2)orA-B=\frac{A^2-B^2}{A^2+AB+B^2}]$
$={lim}_{h\to 0}\left[\frac{8-(8+h)}{h\left\{23\sqrt{8+h}\right\}}\{4+2(8+h{)}^{\frac{1}{2}}+(8+h{)}^{\frac{2}{2}}\}\right]$
$=\left[\frac{-1}{2\times 3\sqrt{8}\{4+2\times 8^{\frac{1}{2}}+8^{\frac{2}{3}}\}}\right]$
$=\frac{-1}{2\times 2(4+4+4)}$
$=\frac{-1}{48}$