What is the next limit equal to? $$\lim_{n\to\infty}(\sqrt{n^2+n}-\sqrt[3]{n^3+1})=?$$
I tried amplifying the hole substraction to form the formula $$a^3-b^3$$ but didn't worked out. Can you help me figure it out?
 A: Generalized binomial theorm:
$(a+b)^k = a^k + ka^{k-1}b + \frac {k(k-1)}{2} a^{k-2}b^2 + \cdots$
$(n^2+n)^\frac 12 = n + \frac 12  - \frac {1}{8} n^{-1} + \cdots\\
(n^3+1)^\frac 13 = n + \frac 13 n^{-2} + \cdots$
Subtract one from the other and let $n$ go to infinity...
$\frac 12$
A: Note that\begin{align}\sqrt{n^2+n}-\sqrt[3]{n^3+1}&=\sqrt[6]{(n^2+n)^3}-\sqrt[6]{(n^3+1)^2}\\&=\frac{(n^2+n)^3-(n^3+1)^2}{\sqrt[6]{(n^2+n)^3}^5+\sqrt[6]{(n^2+n)^3}^4\sqrt[6]{(n^3+1)^2}+\cdots+\sqrt[6]{(n^3+1)^2}^5}\\&=\frac{3n^5+3n^4-n^3-1}{\sqrt[6]{(n^2+n)^3}^5+\sqrt[6]{(n^2+n)^3}^4\sqrt[6]{(n^3+1)^2}+\cdots+\sqrt[6]{(n^3+1)^2}^5}.\end{align}This quotient behaves as $\dfrac{3n^5}{6n^5}$ and therefore its limit is $\dfrac12$.
A: The key here is the (not so) popular standard limit $$\lim_{x\to a} \frac{x^r-a^r} {x-a} =ra^{r-1}\tag{1}$$ The given expression (the one whose limit is to be evaluated) can be written as $$\dfrac{\sqrt{1+\dfrac{1}{n}}-\sqrt[3]{1+\dfrac{1}{n^3}}}{\dfrac{1}{n}}$$ Now we add and subtract $1 $ in the numerator and see that the expression can be written as $$\dfrac{\sqrt{1+\dfrac{1}{n}}-1}{\dfrac{1}{n}}-\frac{1}{n^2}\cdot\dfrac{\sqrt[3]{1+\dfrac{1}{n^3}}-1}{\dfrac{1}{n^3}}$$ Next we put $x=1+(1/n),y=1+(1/n^3)$ so that $x\to 1,y\to 1$ and the expression can be written as $$\frac{x^{1/2}-1}{x-1}-\frac{1}{n^2}\cdot\frac{y^{1/3}-1}{y-1}$$ and using $(1) $ the desired limit is easily seen to be $$\frac{1}{2}-0\cdot\frac{1}{3}=\frac{1}{2}$$
A: Let $1/n=h$ 
$$\implies\lim_{n\to\infty}(\sqrt{n^2+n}-\sqrt[3]{n^3+1})=\lim_{h\to0^+}\dfrac{\sqrt{1+h}-\sqrt[3]{1+h^3}}h$$
$$=\lim_{h\to0^+}\dfrac{\sqrt{1+h}-1}h-\lim_{h\to0^+}\dfrac{\sqrt[3]{1+h^3}-1}{h^3}\cdot\lim_{h\to0^+}h^2$$
For the first choose 
$\sqrt{1+h}-1=u\implies h=(u+1)^2-1$
and for the second $\sqrt[3]{1+h^3}-1=v\implies h^3=(1+v)^3-1$
