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

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(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

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

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(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,

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