Please help with this limit of a sequence: $\lim\limits_{n\to\infty} \bigg(n\sqrt[3]{1+\frac1n} - n\sqrt[3]{1+\frac1{n^2}}\bigg)$ I have this limit of a sequence: $$\lim_{n\to\infty} \bigg(n\sqrt[3]{1+\frac1n} - n\sqrt[3]{1+\frac1{n^2}}\bigg).$$
Can I modify this expression to this expression by knowing that fractions goes to $0$?: $\lim n-n$. How to modify this solution to get the limit?
 A: Let $a=\sqrt[3]{n^3+n^2},\,b=\sqrt[3]{n^3+n}$ so $a-b=\frac{a^3-b^3}{a^2+ab+b^2}=\frac{n^2-n}{a^2+ab+b^2}\approx\frac{n^2}{n^2+n^2+n^2}=\frac13$.
A: HINT
By binomial approximation
$$\sqrt[3]{1+\frac1n}=1+\frac1{3n}+o\left(\frac1n\right)$$
$$\sqrt[3]{1+\frac1{n^2}}=1+\frac1{3n^2}+o\left(\frac1{n^2}\right)$$
these can be obtained by
$$f(x)=\sqrt[3]{1+x} \implies f'(x)=\frac13\frac1{(1+x)^\frac23}\implies f'(0)=\frac13$$
and therefore as $h\to 0$
$$f(h)=f(0)+f'(0)\cdot h+o(h)=1+\frac h3+o(h) $$
A: If $x = (1+1/n)^{\frac{1}{3}}$ and $y = (1+1/n^{2})^{\frac{1}{3}}$, we can multiply and divide the argument by $x^{2}+xy+y^{2}$ to use $x^{3}-y^{3}  =(x-y)(x^{2}+xy+y^{2})$.
Note that $$n\bigg{[}\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{1}{3}}-\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{1}{3}}\bigg{]}\frac{\bigg{[}\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{2}{3}}+\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{1}{3}}\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{1}{3}}+\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{2}{3}}\bigg{]}}{\bigg{[}\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{2}{3}}+\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{1}{3}}\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{1}{3}}+\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{2}{3}}\bigg{]}} \\ =\frac{n\bigg{[}\bigg{(}1+\frac{1}{n}\bigg{)}-\bigg{(}1+\frac{1}{n^{2}}\bigg{)}\bigg{]}}{\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{2}{3}}+\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{1}{3}}\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{1}{3}}+\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{2}{3}}} \\ = \frac{1-\frac{1}{n}}{\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{2}{3}}+\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{1}{3}}\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{1}{3}}+\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{2}{3}}}$$
Now, when $n \to \infty$ the denominator goes to 3 and the numerator goes to 1, so your limit is 1/3.
A: MVT : 
Let $x>y$:  Then $f(x)-(y)=f'(s)(x-y)$, $y<s<x$;
$f(x)=x^{1/3}$.
$n((1+1/n)^{1/3}-(1+1/n^2)^{1/3})=$
$n(1/3)(1+t)^{-2/3}(1/n-1/n^2)$, where $1+1/n^2 <t<1+1/n$;
$\lim_{n \rightarrow \infty} t=0$.
$(\lim_{n \rightarrow \infty}(1/3)(1+t)^{-2/3})\cdot $
$(\lim_{ n \rightarrow \infty}(n(1/n-1/n^2))=$
$(1/3) \cdot 1$.
