$\underset{h→0}{\lim}\begin{Bmatrix}\frac{1}{h(8+h)^{1/3}}-\frac{1}{2h}\end{Bmatrix}$ is equal to:
Answer & explanation
Correct answer: option 4
$\underset{h→0}{\lim}\begin{Bmatrix}\frac{2-(h+8)^{1/3}}{2h(h+8)^{1/3}}\end{Bmatrix}=\underset{h→0}{\lim}\begin{Bmatrix}\frac{[8^{1/3}-(h+8)^{2/3}]+[8^{2/3}+(h+8)^{2/3}+8^{1/3}(h+8)^{1/3}]}{2h(h+8)^{1/3}[8^{2/3}+(h+8)^{2/3}+8^{1/3}(h+8)^{1/3}]}\end{Bmatrix}$
$=\underset{h→0}{\lim}\begin{Bmatrix}\frac{[8-(h+8)]}{2h(h+8)^{1/3}[4+(→4)+(→2)(→2)]}=\frac{-1}{2(→2)(→2)}=\frac{-1}{48}\end{Bmatrix}$
$(Using:[x-y]=(x^{1/3}-y^{1/3}(x^{2/3}+y^{2/3}+(xy)^{1/3}))$