Question

Help with this lim when h--> 0

Answer 1

\[\frac{ \sqrt[3]{h+2}-\sqrt[3]{2} }{ h } \]

Answer 2

can you use the definition of derivative ?

Answer 3

\[\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}=\frac{d}{dx}f(x)\]

Answer 4

\[\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h}=\frac{d}{dx}f(x)|_{x=a}\]

Answer 5

freckles' suggestion is a good way of showing your understanding of the derivative's definition, but in case you're actually computing the derivative of \(f\) from the definition, you can tackle this by multiplying by
\[\frac{\sqrt[3]{(h+2)^2}+\sqrt[3]{h+2}\sqrt[3]{2}+\sqrt[3]{2^2}}{\sqrt[3]{(h+2)^2}+\sqrt[3]{h+2}\sqrt[3]{2}+\sqrt[3]{2^2}}\]
Why? Notice that the numerators cancel nicely:
\[\left(\sqrt[3]{h+2}-\sqrt[3]2\right)\left(\sqrt[3]{(h+2)^2}+\sqrt[3]{h+2}\sqrt[3]{2}+\sqrt[3]{2^2}\right)=(h+2)-2=h\]which means your original limit is equivalent to
\[\lim_{h\to0}\frac{h}{h\left(\sqrt[3]{(h+2)^2}+\sqrt[3]{h+2}\sqrt[3]{2}+\sqrt[3]{2^2}\right)}\\\quad\quad\quad\quad=\lim_{h\to0}\frac{1}{\sqrt[3]{(h+2)^2}+\sqrt[3]{h+2}\sqrt[3]{2}+\sqrt[3]{2^2}}\]