Evaluate $\lim\limits_{x\to+\infty}\frac{\left(x+\sqrt{x^2-1}\right)^n+\left(x-\sqrt{x^2-1}\right)^n}{x^n},n\in \mathbb{N}$

Question

If we use the following $$a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}\right)=u\times t$$

$$u=x+\sqrt{x^2-1}-x+\sqrt{x^2-1}=2\sqrt{x^2-1}=2x\sqrt{1-\frac{1}{x}}$$

Now, the limit is $$2\lim\limits_{x\to+\infty}\frac{x\sqrt{1-\frac{1}{x}}\times t}{x^n}$$

What to do next, if this is a good approach?

Answer 1

Use this properties: $$\frac{a^n}{b^n}=\left(\frac{a}{b}\right)^n\tag{1}$$ $$\lim_{x\to\infty}\left[f(x)\pm g(x)\right]=\lim_{x\to\infty}f(x)\pm \lim_{x\to\infty}g(x)\tag{2}$$

if and only if the limit of $f(x)$ and $g(x)$ exists

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Therefore:

$$\begin{align}\lim_{x\to\infty}\frac{(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n}{x^n}&=\lim_{x\to\infty}\left(\frac{x+\sqrt{x^2-1}}{x}\right)^n+\lim_{x\to\infty}\left(\frac{x-\sqrt{x^2-1}}{x}\right)^n\\&=\lim_{x\to\infty}\left[\frac{x(1+\sqrt{1-\frac{1}{x^2}})}{x}\right]^n+\lim_{x\to\infty}\left[\frac{x(1-\sqrt{1-\frac{1}{x^2}}}{x}\right]^n\\&=\underbrace{\lim_{x\to\infty}\left(1+\sqrt{1-\frac{1}{x^2}}\right)^n}_{2^n}+\underbrace{\lim_{x\to\infty}\left(1-\sqrt{1-\frac{1}{x^2}}\right)^n}_{0}\\&=2^n\end{align}$$

Answer 2

Correction to your answer: $$2\sqrt{x^2-1}=2x\sqrt{1-\frac{1}{x^2}}$$

Your approach is going to have trouble because $t$ is a fairly complicated mess.

Note:

$$\frac{x+\sqrt{x^2-1}}{x} = 1+\sqrt{1-\frac{1}{x^2}}\to 2\text{ as } x\to +\infty$$

and

$$\frac{x-\sqrt{x^2-1}}{x}= 1-\sqrt{1-\frac{1}{x^2}}\to 0\text{ as }x\to+\infty$$

Answer 3

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

Answer 4

Let $a= x+\sqrt{x^2-1}$, then we have $x-\sqrt{x^2-1}=\frac{1}{a}$ and $a+ \frac{1}{a}= 2x$. Next, we can write $$\lim\limits_{x\to+\infty}\frac{(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n}{x^n} =\lim\limits_{a\to+\infty}\frac{a^n+\frac{1}{a^n}}{\frac{1}{2^n}(a+\frac{1}{a})^n}=\lim\limits_{a\to+\infty}\frac{a^n}{\frac{1}{2^n}(a)^n}= 2^{n}$$