lim _{n} {rightarrow infty} sqrt{2} left[ fra | vedclass

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Question

(D) $n \rightarrow \infty$ के लिए: चूंकि $2-\sqrt{2} < 1$,इसलिए $\lim _{n \rightarrow \infty} (2-\sqrt{2})^n = 0$. अंश और हर को $(2+\sqrt{2})^n$ से वि

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(D) $n \rightarrow \infty$ के लिए: चूंकि $2-\sqrt{2} अंश और हर को $(2+\sqrt{2})^n$ से विभाजित करने पर: $= \lim _{n}$ ${\rightarrow \infty} \sqrt{2} \left[ \frac{1 + \left( \frac{2-\sqrt{2}}{2+\sqrt{2}} \right)^n}{1 - \left( \frac{2-\sqrt{2}}{2+\sqrt{2}} \right)^n} \right] = \sqrt{2} \left[ \frac{1+0}{1-0} \right] = \sqrt{2}$.

$\lim _{n}$ ${\rightarrow \infty} \sqrt{2} \left[ \frac{(2+\sqrt{2})^n + (2-\sqrt{2})^n}{(2+\sqrt{2})^n - (2-\sqrt{2})^n} \right] =$

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