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

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 $\frac{1}{3}$.
