Question

lim h-->0 (27+h)^1/3 - 3/h

Answer 1

Which function will be the most powerful? Do you know

Answer 2

Substitute \(a=(27+h)^{1/3}\) and \(b=27^{1/3}=3\). Then,
\[\begin{align*}\lim_{h\to0}\frac{(27+h)^{1/3}-3}{h}&=\lim_{h\to0}\frac{a-b}{h}\\
&=\lim_{h\to0}\frac{a-b}{h}\cdot\frac{a^2+ab+b^2}{a^2+ab+b^2}\\
&=\lim_{h\to0}\frac{a^3-b^3}{h\left(a^2+ab+b^2\right)}\\
&=\lim_{h\to0}\frac{\left((27+h)^{1/3}\right)^3-\left((27)^{1/3}\right)^3}{h\left(\left((27+h)^{1/3}\right)^2+(27+h)^{1/3}27^{1/3}+\left(27^{1/3}\right)^2\right)}\\
&=\lim_{h\to0}\frac{27+h-27}{h\left((27+h)^{2/3}+(27+h)^{1/3}27^{1/3}+27^{2/3}\right)}\\
&=\lim_{h\to0}\frac{1}{(27+h)^{2/3}+(27+h)^{1/3}27^{1/3}+27^{2/3}}\\
&=\frac{1}{27^{2/3}+27^{1/3}27^{1/3}+27^{2/3}}\\
&=\frac{1}{3\left(27^{2/3}\right)}
\end{align*}\]

Answer 3

Also, \(27^{2/3}=\left(27^{1/3}\right)^2=3^2=9\).

Answer 4

Wow that's a weird fancy way to do that sith :o

I think for this problem they just want you to recognize that the limit represents a derivative.\[\Large\bf\sf f'(27)\quad=\quad \lim_{h\to0}\frac{(27+h)^{1/3}-(27)^{1/3}}{h}\]Where,\[\Large\bf\sf f(x)\quad=\quad x^{1/3}\]