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Evaluate $$\lim_{n \to \infty} \frac{(1+\sqrt 2)^n+(1-\sqrt 2)^n}{(1+\sqrt 2)^n-(1-\sqrt 2)^n}.$$

I tried to expand using Newton's Binomial Theorem, but it didn't work.

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    $\begingroup$ Divide throughout by $(1+\sqrt{2})^n$ and observe that $\left|\dfrac{1-\sqrt{2}}{1+\sqrt{2}}\right| < 1$. $\endgroup$ – Muralidharan Nov 15 '18 at 16:22
  • $\begingroup$ @Muralidharan you might post your comment as an answer :) $\endgroup$ – Nosrati Nov 15 '18 at 16:27
  • $\begingroup$ observe that $|1-\sqrt{2}|<1$ $\endgroup$ – Vasya Nov 15 '18 at 16:51
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$\lim_{n \to \infty} \frac{(1+\sqrt 2)^n+(1-\sqrt 2)^n}{(1+\sqrt 2)^n-(1-\sqrt 2)^n}$ $= \lim_{n\to \infty }\frac{1+a^n}{1-a^n} $ such that $a=\frac{(1-2^{\frac{1}{2}})}{(1+2^{\frac{1}{2}})} $ , $|a|<1$. So lim is equal to $\frac{1-0}{1+0}=1$

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    $\begingroup$ Dadrahm.Maybe you could add |a| <1. $\endgroup$ – Peter Szilas Nov 15 '18 at 16:49
  • $\begingroup$ Thank you . . . $\endgroup$ – Darman Nov 15 '18 at 17:43
  • $\begingroup$ Dahdram.A pleasure +. $\endgroup$ – Peter Szilas Nov 15 '18 at 17:55
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Roughly, $|1-\sqrt 2| \lt 1$, so a high power of it will go to $0$. $1+\sqrt 2 \gt 1$, so a high power of it will be large and positive. We can ignore the two small terms and be left with the fixed ratio $1$. Depending on what theorems you have proved about limits that may be enough.

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We have that

  • $|1-\sqrt 2|<1 \implies (1-\sqrt 2)^n \to 0$

therefore

$$\frac{(1+\sqrt 2)^n+(1-\sqrt 2)^n}{(1+\sqrt 2)^n-(1-\sqrt 2)^n}\sim \frac{(1+\sqrt 2)^n}{(1+\sqrt 2)^n}=1$$

or more rigoursly

$$\frac{(1+\sqrt 2)^n+(1-\sqrt 2)^n}{(1+\sqrt 2)^n-(1-\sqrt 2)^n}= \frac{(1+\sqrt 2)^n}{(1+\sqrt 2)^n}\frac{1+\frac{(1-\sqrt 2)^n}{(1+\sqrt 2)^n}}{1-\frac{(1-\sqrt 2)^n}{(1+\sqrt 2)^n}}\to \frac{1+0}{1-0}$$

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