this is not a homework question.
how do i reach the limit of:
i tried taking out $n^3$ and then i got $n-n$ which is $0$, but the true answer is $2/3$.
i can't understand why the answer is $2/3$ and what method to use here.
thank you very much in advance,
yaron.
On
Perhaps easier to simplify first. We want to get rid of the cube roots, so remember that $a^3-b^3 = (a-b) \cdot \left(a^2 + ab + b^2\right)$. Note that
$$ \begin{split} \sqrt[3]{n^3+2n^2+4} - \sqrt[3]{n^3+1} &= \frac{\left(\sqrt[3]{n^3+2n^2+4} - \sqrt[3]{n^3+1}\right) \left( {(n^3+2n^2+4)^{2/3} + (n^3+1)^{2/3} + \sqrt[3]{(n^3+2n^2+4)(n^3+1)}} \right)} {(n^3+2n^2+4)^{2/3} + (n^3+1)^{2/3} + \sqrt[3]{(n^3+2n^2+4)(n^3+1)}} \\ &= \frac{n^3+2n^2+4 - n^3 - 1} {(n^3+2n^2+4)^{2/3} + (n^3+1)^{2/3} + \sqrt[3]{(n^3+2n^2+4)(n^3+1)}} \\ &= \frac{2n^2+3} {(n^3+2n^2+4)^{2/3} + (n^3+1)^{2/3} + \sqrt[3]{(n^3+2n^2+4)(n^3+1)}} \\ \end{split} $$ Divide through by $n^2$ and with $n \to \infty$ this will look like $\frac{2}{1+1+1} = 2/3$.
When you pick out $n^3$, you should get $$-n(1+\frac{1}{n^3})^{1/3}+n(1+\frac{2}{n}+\frac{4}{n^3})^{1/3}$$
The cube-root of $1+x$ is close to $1+(x/3)$ if $x$ is small, so your expression is close to $$-n(1+\frac{1}{3n^3})+n\left(1+\frac{2}{3n}+\frac{4}{3n^3}\right)$$ As you can see, the largest term that remains is n(2/3n) = 2/3