# Help Proving a Pair of Simple Limits

I would appreciate someone showing me the best way to algebraically prove a pair of related limit problems. I intuitively see that they are true since $\sqrt{1+4h^2}\approx2h$ but if someone could show me how to flesh it out that would be great. Here they are. For $h>0$:

$$\begin{split} &\lim_{n\rightarrow\infty}(-2h+\sqrt{1+4h^2})^n=0 \\ &\lim_{n\rightarrow\infty}\left|(-2h-\sqrt{1+4h^2})^n\right|=\infty \end{split}$$

I realize this isn't as interesting and rewarding a problem as most on this site but any help would be much appreciated. Thanks

One way of solving this, trying to use your intuition that $\sqrt{1+4h^2}\approx 2h$ goes as follows. One has that $\sqrt{1+4h^2}> \sqrt{4h^2}=2h$, and that $\sqrt{1+4h^2} < \sqrt{1+4h+4h^2}=\sqrt{(1+2h)^2}=1+2h$.
For the first limit this gives $-2h+\sqrt{1+4h^2}< -2h+2h+1=1$ and $-2h+\sqrt{1+4h^2}>-2h+2h=0$. it is known that $\lim_{n\to\infty}a^n = 0$ for $|a|<1$, so the first limit follows.
For the second limit, note that $2h+\sqrt{1+4h^2}>\sqrt{1+4h^2}>\sqrt{1}=1$. And as $\lim_{x\to\infty}|a|^n=\infty$ if $|a|>1$ the second limit follows.
Hint: The second one is easy: use $|a^n| = |a|^n$. For the first one: use $$A-B = \frac{A^2 - B^2}{A+B}$$