MCQ
Let $x_n=\left(2^n+3^n\right)^{1 / 2 n}$ for all natural numbers $n$. Then,
  • A
    $\lim _{n \rightarrow \infty} x_n=\infty$
  • $\lim _{n \rightarrow \infty} x_n=\sqrt{3}$
  • C
    $\lim _{n \rightarrow \infty} x_n=\sqrt{3}+\sqrt{2}$
  • D
    $\lim _{n \rightarrow \infty} x_n=\sqrt{5}$

Answer

Correct option: B.
$\lim _{n \rightarrow \infty} x_n=\sqrt{3}$
b
(b)

We have,

$x_n =\left(2^n+3^n\right)^{\frac{1}{2 n}}$

$\Rightarrow \quad \lim _{n \rightarrow \infty} x_n =\lim _{n \rightarrow \infty}\left(2^n+3^n\right)^{\frac{1}{2 n}}$

$=\lim _{n \rightarrow \infty} 3^{n / 2 n}\left[\left(\frac{2}{3}\right)^n+1\right]^{\frac{1}{2 n}}$

$=3^{3 / 2} \times 1=\sqrt{3}$

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