<s>[INST] 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?
 [/INST] 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 $$\frac{1}{2}-0\cdot\frac{1}{3}=\frac{1}{2}$$
</s>