# How do you find the limit lim_(x->0)((8+h)^(1/3)-2)/h without using L'Hopital's Rule?

## ${\lim}_{x \to 0} \frac{{\left(8 + h\right)}^{\frac{1}{3}} - 2}{h}$

Jul 13, 2018

Answer to this question is $\frac{1}{12}$.

#### Explanation:

$L i {m}_{x \to 0} \frac{{\left(8 + h\right)}^{\frac{1}{3}} - 2}{h}$

Now notice that we need to get rid of exponent $\frac{1}{3}$. Therefore we multiply the numerator and denominator by an expression that will make it free of exponent $\frac{1}{3}$.

We know that: ${A}^{3} - {B}^{3} = \left(A - B\right) \left({A}^{2} + A B + {B}^{2}\right)$
In our case $A = {\left(8 + h\right)}^{\frac{1}{3}}$ and $B = 2$. I notice that if I multiply numerator and denominator by $\left({A}^{2} + A B + {B}^{2}\right)$ i.e., $\left({\left(8 + h\right)}^{\frac{2}{3}} + 2 {\left(8 + h\right)}^{\frac{1}{3}} + 4\right)$, then I will get ${A}^{3} = {\left({\left(8 + h\right)}^{\frac{1}{3}}\right)}^{3} = \left(8 + h\right)$. and hence A becomes free of exponent $\frac{1}{3}$.

Multiply and divide by $\left({\left(8 + h\right)}^{\frac{2}{3}} + 2 {\left(8 + h\right)}^{\frac{1}{3}} + 4\right)$

$\implies L i {m}_{x \to 0} \left(\frac{{\left(8 + h\right)}^{\frac{1}{3}} - 2}{h} \cdot \frac{{\left(8 + h\right)}^{\frac{2}{3}} + 2 {\left(8 + h\right)}^{\frac{1}{3}} + 4}{{\left(8 + h\right)}^{\frac{2}{3}} + 2 {\left(8 + h\right)}^{\frac{1}{3}} + 4}\right)$

$\implies L i {m}_{x \to 0} \frac{{\left(8 + h\right)}^{\frac{3}{3}} - {2}^{3}}{h \cdot \left({\left(8 + h\right)}^{\frac{2}{3}} + 2 {\left(8 + h\right)}^{\frac{1}{3}} + 4\right)}$

$\implies L i {m}_{x \to 0} \frac{8 + h - 8}{h \cdot \left({\left(8 + h\right)}^{\frac{2}{3}} + 2 {\left(8 + h\right)}^{\frac{1}{3}} + 4\right)}$

$\implies L i {m}_{x \to 0} \frac{h}{h \cdot \left({\left(8 + h\right)}^{\frac{2}{3}} + 2 {\left(8 + h\right)}^{\frac{1}{3}} + 4\right)}$

$\implies L i {m}_{x \to 0} \frac{\textcolor{red}{\cancel{h}}}{\textcolor{red}{\cancel{h}} \cdot \left({\left(8 + h\right)}^{\frac{2}{3}} + 2 {\left(8 + h\right)}^{\frac{1}{3}} + 4\right)}$

$\implies L i {m}_{x \to 0} \frac{1}{{\left(8 + h\right)}^{\frac{2}{3}} + 2 {\left(8 + h\right)}^{\frac{1}{3}} + 4}$

$\implies \frac{1}{{\left(8 + 0\right)}^{\frac{2}{3}} + 2 {\left(8 + 0\right)}^{\frac{1}{3}} + 4}$

$\implies \frac{1}{4 + 2 \cdot 2 + 4}$

$\implies \frac{1}{12}$