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

The Correct Option is C

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.