For each rational $\alpha$ find $$\lim_{n \to \infty} { \sqrt{n^2 +n^{\alpha}} } - n$$ Only elementary inequalities are allowed
Hint: Look at $\alpha < 1, 1 \le \alpha < 2, \alpha \ge 2$
By rationalization, we have $$\lim_{n \to \infty} { \sqrt{n^2 +n^{\alpha}} } - n=\lim_{n \to \infty} \frac{n^{\alpha-1}}{ \sqrt{1+n^{\alpha-2}}+1}.$$ From here it is clear that for $\alpha \ge 2$ the expressions $\to \infty$ and for $ \alpha =1 $ the limit =1 and as for $ \alpha \lt 1$ the limit is $0$. I am stuck for the case $1\leq \alpha <2$.
Elementary approach: for every $\alpha$ in $(1,2)$, $$n^{\alpha-2}\leqslant1\qquad\text{and}\qquad n^{\alpha-1}\to+\infty$$ hence $$\frac{n^{\alpha-1}}{\sqrt{1+n^{\alpha-2}}+1}\geqslant\frac{n^{\alpha-1}}{\sqrt{1+1}+1}\geqslant\frac{n^{\alpha-1}}3\to+\infty$$