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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{equation}\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}\end{equation}

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

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2 Answers 2

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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.

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  • $\begingroup$ Ah. Completing the square to show that the radical expression is less than 1+2h is very good. Thanks $\endgroup$ Oct 26, 2014 at 2:21
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Hint: The second one is easy: use $|a^n| = |a|^n$. For the first one: use $$A-B = \frac{A^2 - B^2}{A+B}$$

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  • $\begingroup$ I like that. That is a quick trick. Thanks $\endgroup$ Oct 26, 2014 at 2:19

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