Limits #2

$n( \sqrt[3]{ \frac{1}{n} - 1} + n)$

$= n\Big(( \frac{1}{n} - 1 )^{\frac{1}{3}} + 1^{\frac{1}{3}})$

now $a + b =\frac{ a^{2} + b^{3}}{a^{2} - ab +b^{2}}$

thus

$n ( \frac{( \frac{1}{n} - 1) + 1}{ ( \frac{1}{n} - 1)^{\frac{2}{3}} - (\frac{1}{n} - 1)^{\frac{1}{3}} + 1}) = \frac{1}{3}$

$\displaystyle lim_{n \to \infty } (n^{2} - n^{3})^{\frac{1}{3}} + n$

Now take $-n^{3}$ common from root

$\displaystyle -n(1 - \frac{1}{n})^{\frac{1}{3}} + n$

Now as $n \to \infty$

$\frac{1}{n} \to 0$

Binomial says

$\displaystyle (1 + x)^{n} = 1 + nx$ (when x <<< 1)

$\displaystyle \lim_{n \to \infty} -n( 1 - \frac{1}{3n}) + n = -n + \frac{1}{3} + n = \boxed{\frac{1}{3}}$

$\begin{aligned} L & = \lim_{n \to \infty} \left(\sqrt[3]{n^2-n^3} + n \right) \\ & = \lim_{n \to \infty} n \left(\sqrt[3]{\frac 1n -1} + 1 \right) \\ & = \lim_{n \to \infty} n \left(1 - {\color{#3D99F6} \sqrt[3]{1- \frac 1n}} \right) & \small \color{#3D99F6} \text{By Taylor's series expansion} \\ & = \lim_{n \to \infty} n \left(1 - {\color{#3D99F6} \left(1- \frac 1{3n} - \frac 1{9n^2} - \frac 5{81n^3} - \cdots \right)} \right) \\ & = \lim_{n \to \infty} \left( \frac 13 + \frac 1{9n} + \frac 5{81n^2} + \cdots \right) \\ & = \frac 13 \end{aligned}$

Therefore, $a+b=1+3=\boxed{4}$ .

$\begin{aligned} L &= \displaystyle \lim_{n \to \infty} \sqrt[3]{n^2- n^3} + n \\ &= \lim_{n \to \infty} n \sqrt[3]{\dfrac1n -1} + n \\ &= \lim_{n \to \infty} n \left( \sqrt[3]{\dfrac1n -1} + 1 \right) \\ &= \lim_{n \to \infty} \dfrac{\sqrt[3]{\frac1n -1} + 1}{\frac1n} & \small\color{#3D99F6}{\dfrac00 \text{ case so apply L'Hopital's Rule}} \\ &= \lim_{n \to \infty} \dfrac 13 \cdot {\left( \dfrac1n -1 \right)}^{{-2}/{3}} \\ &= \boxed{\dfrac 13} \end{aligned}$

In getting the closest value (though not exactly), I let infinity be equal to 999999.. Upon substituting, I got the answer 0.333... which is equal to 1/3.. hence, 1+3=4..