Question about $\lim_{n\rightarrow\infty} (\sqrt{n} - \sqrt{n - n^c})$ and $\lim_{n\rightarrow\infty} (\sqrt[3]{n + n^c} - \sqrt[3]{n})$

Question

I have following exercises:

1) Find $\lim_{n\rightarrow\infty} (\sqrt{n} - \sqrt{n - n^c})$, $0\leq c\leq 1$
hint: make a case distinction for $c < \frac{1}{2}, c = \frac{1}{2}, c > \frac{1}{2}$

2) Find $\lim_{n\rightarrow\infty} (\sqrt[3]{n + n^c} - \sqrt[3]{n})$, $0\leq c\leq 1$
hint: make a case distinction for $c < \frac{2}{3}, c = \frac{2}{3}, c > \frac{2}{3}$

My question: without the hints, how could I derive the values $\frac{1}{2}$ and $\frac{2}{3}$ on my own?

Answer 1

I like to use series for these. I will do the second one.
Cases $c=0,1$ are trivial, so assume $0<c<1$. $$ \sqrt[3]{n+n^c} = (n+n^c)^{1/3} = n^{1/3}(1+n^{c-1})^{1/3} $$ But $n^{c-1} \to 0$, so (Taylor series or Newton's binomial series) $$ (1+n^{c-1})^{1/3} = 1+\frac{1}{3} n^{c-1}-\frac{1}{9} n^{2c-2}+\dots \\ \sqrt[3]{n+n^c} = n^{1/3}+\frac{1}{3} n^{c-2/3}-\frac{1}{9} n^{2c-5/3}+\dots \\ \sqrt[3]{n+n^c} - \sqrt[3]{n} = \frac{1}{3} n^{c-2/3}-\frac{1}{9} n^{2c-5/3}+\dots $$ Now we can see where the $2/3$ comes from. If $c>2/3$, then the leading term has a positive power on $n$, so we get limit $\infty$. If $c<2/3$ then all terms have negative exponent, so we get limit $0$. And if $c=2/3$, the leading term is $\frac{1}{3}$ and all remaining terms have negatvie exponent, so we get limit $\frac{1}{3}$.

Answer 2

$$\sqrt{n}-\sqrt{n-n^c}=\frac{n^c}{\sqrt{n}+\sqrt{n-n^c}}=\frac1{\sqrt{n^{\color{blue}{1-2c}}}+\sqrt{n^{1-2c}-n^{-c}}}$$

\begin{align}\sqrt[3]{n+n^c}-\sqrt[3]{n}&=\frac{n^c}{\sqrt[3]{(n+n^c)^2}+\sqrt[3]{n^2}+\sqrt[3]{n(n+n^c)}}\\&=\frac{1}{\sqrt[3]{(n+n^c)^2n^{-3c}}+\sqrt[3]{n^{\color{blue}{2-3c}}}+\sqrt[3]{n^{1-3c}(n+n^c)}} \end{align}