# How to evaluate $\lim_{n \to \infty} \frac{(1+\sqrt 2)^n+(1-\sqrt 2)^n}{(1+\sqrt 2)^n-(1-\sqrt 2)^n}$?

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.

• Divide throughout by $(1+\sqrt{2})^n$ and observe that $\left|\dfrac{1-\sqrt{2}}{1+\sqrt{2}}\right| < 1$. – Muralidharan Nov 15 '18 at 16:22
• @Muralidharan you might post your comment as an answer :) – Nosrati Nov 15 '18 at 16:27
• observe that $|1-\sqrt{2}|<1$ – Vasya Nov 15 '18 at 16:51

$$\lim_{n \to \infty} \frac{(1+\sqrt 2)^n+(1-\sqrt 2)^n}{(1+\sqrt 2)^n-(1-\sqrt 2)^n}$$ $$= \lim_{n\to \infty }\frac{1+a^n}{1-a^n}$$ such that $$a=\frac{(1-2^{\frac{1}{2}})}{(1+2^{\frac{1}{2}})}$$ , $$|a|<1$$. So lim is equal to $$\frac{1-0}{1+0}=1$$

• Dadrahm.Maybe you could add |a| <1. – Peter Szilas Nov 15 '18 at 16:49
• Thank you . . . – Darman Nov 15 '18 at 17:43
• Dahdram.A pleasure +. – Peter Szilas Nov 15 '18 at 17:55

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.

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

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