If $(1+\sqrt{2})^n=a_{n}+b_{n}\sqrt{2}\;,(\forall n\in \mathbb{N}).$Then $\lim_{n\rightarrow \infty}\frac{a_{n}}{b_{n}} = $

By the binomial theorem,

$$(1+\sqrt2)^n=a_n+b_n\sqrt2\implies(1-\sqrt2)^n=a_n-b_n\sqrt2.$$

Then

$$\frac{a_n}{b_n}=\sqrt2\frac{(1+\sqrt2)^n+(1-\sqrt2)^n}{(1+\sqrt2)^n-(1-\sqrt2)^n}=\sqrt2\frac{1+\left(\dfrac{1-\sqrt2}{1+\sqrt2}\right)^n}{1-\left(\dfrac{1-\sqrt2}{1+\sqrt2}\right)^n}.$$

This gives linearly converging rational approximations of $\sqrt2$. For instance, with $n=10$,

$$\sqrt2\approx\frac{6726}{4756}$$ with a relative error on the order of

$$2\left(\frac{1-\sqrt2}{1+\sqrt2}\right)^{10}\approx4.4\cdot10^{-8}.$$

Here is another path.

$$(a_n+b_n\sqrt{2})(1+\sqrt{2})=a_n+2b_n+(a_n+b_n)\sqrt{2}$$

thus if we let $c_n=\frac{a_n}{b_n}$ then

$$c_{n+1}=1+\frac{1}{c_n+1}$$

One can check by induction that the odd terms are increasing, the even decreasing and their difference goes to zero. Thus the limit must satisfy

$$L=1+\frac{1}{L+1}$$ and so $L=\sqrt{2}$.

Note first that the recursion

$$a_{n+1}+b_{n+1}\sqrt2=(1+\sqrt2)(a_n+b_n\sqrt2)=(a_n+2b_n)+(a_n+b_n)\sqrt2$$

implies $a_1,a_2,a_3,\ldots$ and $b_1,b_2,b_3,\ldots$ are both strictly increasing sequences of positive integers, since $a_1=b_1=1$ and $a_n,b_n\gt0$ implies $a_{n+1}=a_n+2b_n\gt a_n$ and $b_{n+1}=a_n+b_n\gt b_n$. In particular, $b_n\to\infty$ as $n\to\infty$.

Now, by the formal symmetry between $\sqrt2$ and $-\sqrt2$ and the general algebraic identity $(x+y)(x-y)=x^2-y^2$, we have

$$\begin{align} (1+\sqrt2)^n=a_n+b_n\sqrt2&\implies(1-\sqrt2)^n=a_n-b_n\sqrt2\\ &\implies(-1)^n=a_n^2-2b_n^2\\ &\implies{a_n\over b_n}=\sqrt{2\left(1+{(-1)^n\over b_n^2}\right)}\\ &\implies\lim_{n\to\infty}{a_n\over b_n}=\sqrt2 \end{align}$$