Question

Let $l_{n } = \frac{2^{n } + \left(-2\right)^{n} }{2^{n}}$ and $L_{n} = \frac{2^{n} + \left(- 2\right)^{n}}{3^{n}}$ then as $n \to\infty$

• A. Both the sequences have limits
• B. $\displaystyle\lim_{n \to \infty} \, \, l_{n}$ exists but $\displaystyle\lim_{n \to \infty} \, L_{n}$ does not exist
• C. $\displaystyle\lim_{n \to \infty} \, \, l_{n}$ does not exist but $\displaystyle\lim_{n \to \infty} \, L_{n}$ exists
• D. Both the sequences do not have limits.

Answer 1

Answer: (C)

$\displaystyle\lim_{n \to \infty} \, \, l_{n}$ does not exist but $\displaystyle\lim_{n \to \infty} \, L_{n}$ exists

Answer 2

Given, ln $=\frac{2^{n}+(-2)^{n}}{2^{n}}=1+\frac{(-2)^{n}}{2^{n}}\,\\...(i)$
And $L_{n}=\frac{2^{n}+(-2)^{2 n}}{3^{n}}\,...(ii)$
Now, from E (i), we get
$\displaystyle\lim _{n \rightarrow \infty}$ ln $=\begin{cases}0, & \text { when } n \text { is odd } \\\ 2, & \text { when } n \text { is even }\end{cases}$
$\therefore\, \displaystyle\lim _{n \rightarrow \infty}$ does not exist
and $\displaystyle\lim _{n \rightarrow \infty} L_{n}=0$
$\therefore \, \displaystyle\lim _{x \rightarrow \infty} L_{n}$ exist.