Let x_n = (2^n + 3^n)^{frac{1}{2n}} for all n | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We have $x_n = (2^n + 3^n)^{\frac{1}{2n}}$.Taking the limit as $n 	o \infty$:$\lim_{n 	o \infty} x_n = \lim_{n 	o \infty} (3^n ((\frac{2}{3})^n

Vedclass · Accepted Answer

$\lim_{n 	o \infty} x_n = \sqrt{3}$

Vedclass · Answer

(B) We have $x_n = (2^n + 3^n)^{\frac{1}{2n}}$.Taking the limit as $n 	o \infty$:$\lim_{n 	o \infty} x_n = \lim_{n 	o \infty} (3^n ((\frac{2}{3})^n + 1))^{\frac{1}{2n}}$$= \lim_{n 	o \infty} (3^n)^{\frac{1}{2n}} \cdot ((\frac{2}{3})^n + 1)^{\frac{1}{2n}}$$= \lim_{n 	o \infty} 3^{\frac{1}{2}} \cdot ((\frac{2}{3})^n + 1)^{\frac{1}{2n}}$Since $\lim_{n 	o \infty} (\frac{2}{3})^n = 0$ and $\lim_{n 	o \infty} \frac{1}{2n} = 0$,the expression becomes:$= \sqrt{3} \cdot (0 + 1)^0 = \sqrt{3} \cdot 1 = \sqrt{3}$.

Let $x_n = (2^n + 3^n)^{\frac{1}{2n}}$ for all natural numbers $n$. Then,

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{a^x} - {b^x}}}{x}} \right) = $

Let $[x]$ denote the greatest integer less than or equal to $x$. Then $\lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right)=$

$\lim _{x \rightarrow-\infty} \log _e(\cosh x)+x=$

$\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}} = $

$\mathop {\lim }\limits_{n \to \infty } \left\{ {\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}} + \frac{3}{{{n^2}}} + \dots + \frac{n}{{{n^2}}}} \right\}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module