Find $\lim_{n\to \infty}$ $n+(n^2-n^3)^{1/3}$.

Question

Visited Help With Find Calculating $\lim_{n \to \infty} (n^3 + n^2)^\frac {1}{3} - (n^3 + n)^\frac {1}{3}$.

Way$1$. This didn't helped me much.I tried to make the binomial expansion, it came up like this:

$(n^3(\frac{1}{n}-1))+n= n(\frac{1}{n}-1)^{1/3}$+n (but I can not take negative common! It will turn out to give $(-1)^{1/3} $ which seems that it is not a correct way to do this question(may be!)).

Way$2.$ The link that I mentioned above (referring to the second answer) after multiplying and dividing I got $\frac {n^2}{(n^2-n^3)^{2/3}+n^2+n^2(n^2-n^3)^{1/3}}$ ( I am getting a wrong answer ).

The correct answer is $1/3$. Any help? Thanks.

Answer 1

Hint

Set $1/n=h$ to get

$$\lim_{h\to0}\dfrac{1+\sqrt[3]{h-1}}h$$

Now set $\sqrt[3]{h-1}=v$ to find $$\lim_{v\to-1}\dfrac{1+v}{1+v^3}=?$$

Answer 2

$$\begin{eqnarray*}n+(n^2-n^3)^{1/3} &=& n-(n^3-n^2)^{1/3}\\ &=& n-n\left(1-\frac{1}{n}\right)^{1/3}\\ &=&n-n\left(1-\frac{1}{3n}+O\left(\frac{1}{n^2}\right)\right)\\&=&\color{red}{\frac{1}{3}}+O\left(\frac{1}{n}\right).\end{eqnarray*}$$

Answer 3

$$ \begin{aligned} \lim_{n\rightarrow \infty}\left[n+(n^2-n^3)^\frac{1}{3}\right] &= \lim_{n\rightarrow \infty}\left[n-\sqrt[3]{(n^3-n^2)}\right] = \lim_{n\rightarrow \infty}n\left[1-\sqrt[3]{1-\frac{1}{n}}\right] = \\ &=\lim_{n\rightarrow \infty}n\frac{\left[1-\sqrt[3]{1-\frac{1}{n}}\right]\left[1^2+1\cdot\sqrt[3]{1-\frac{1}{n}}+\left(\sqrt[3]{1-\frac{1}{n}}\right)^2\right]}{\left[1^2+1\cdot\sqrt[3]{1-\frac{1}{n}}+\left(\sqrt[3]{1-\frac{1}{n}}\right)^2\right]} = \\ &=\left[ \begin{array}{l} \text{numerator can be simplified by the formula}\\ a^3-b^3 = (a-b)(a^2+ab+b^2) \end{array} \right] = \\ &= \lim_{n\rightarrow \infty}\frac{n\left[1^3-\left(\sqrt[3]{1-\frac{1}{n}}\right)^3\right]}{\left[1+\sqrt[3]{1-\frac{1}{n}}+\left(\sqrt[3]{1-\frac{1}{n}}\right)^2\right]} = \\ &= \lim_{n\rightarrow \infty}\frac{\overbrace{n\left[1-\left(1-\frac{1}{n}\right)\right]}^{=1}}{\left[1+\underbrace{\sqrt[3]{1-\frac{1}{n}}}_{\downarrow_{n\rightarrow \infty} \\ 1}+\underbrace{\left(\sqrt[3]{1-\frac{1}{n}}\right)^2}_{\downarrow_{n\rightarrow \infty} \\ 1}\right]} = \frac{1}{3} \end{aligned} $$