Mathematics - Limits Question with Solution | TestHub

We have $a_n+\sqrt{2}{b}_n=(2+\sqrt{2}{)}^n$

$\Rightarrow a_n-\sqrt{2}{b}_n=(2-\sqrt{2}{)}^n$

Therefore $a_n=\frac{1}{2}\left[(2+\sqrt{2}{)}^n+(2-\sqrt{2}{)}^n\right]$ And $b_n=\frac{\left[(2+\sqrt{2}{)}^n-(2-\sqrt{2}{)}^n\right]}{2\sqrt{2}}$

Therefore $\begin{array}{rlr}\frac{a_n}{b_n}& =\sqrt{2}\frac{\left[(2+\sqrt{2}{)}^n+(2-\sqrt{2}{)}^n\right]}{\left[(2+\sqrt{2}{)}^n-(2-\sqrt{2}{)}^n\right]}& =\sqrt{2}\frac{\left[1+{\left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)}^n\right]}{\left[1-{\left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)}^n\right]}\end{array}$

Hence ${\lim }_{n\to \infty }\left(\frac{a_n}{b_n}\right)=\sqrt{2}\left(\frac{1+0}{1-0}\right)\left(\because \frac{2-\sqrt{2}}{2+\sqrt{2}}<1\right)=\sqrt{2}$