$V_{n+1} = \sqrt{V_{n}^2 + (1/5)^n}$
on squaring both side, we get
$V_{n+1}^2 - V_{n}^2 = (1/5)^n$
$V_{2}^2 - V_{1}^2 = (1/5)$
$V_{3}^2 - V_{2}^2 = (1/5)^2$
$V_{4}^2 - V_{3}^2 = (1/5)^3$
$V_{5}^2 - V_{4}^2 = (1/5)^4$
$\vdots$
$V_{n}^2 - V_{n-1}^2 = (1/5)^n-1$
On adding all we get,
$V_{n}^2 - V_{1}^2 = ((1/5) + (1/5)^2 + (1/5)^3 + (1/5)^4 + (1/5)^5 +\dots + (1/5)^{n-1}$
$\quad = ( 1 - (1/5)^{n-1}) / 4$
$\quad = 1/4 - (1/5)^{n-1}/4 $
$\implies V_{n}^2 = V_{1}^2 + 1/4 - (1/5)^{n-1}/4 $
$\quad \quad = 1+ 1/4 - (1/5)^{n-1}/4 $
$\quad \quad = 5/4 - (1/5)^{n-1}/4 $
on taking limit, we get
$V_{n} = \sqrt{5/4}$
$$(v)\lim_{x\to \infty}a^x = \begin{cases}0, &0\leq a < 1 \\1, \quad &a= 1\\\infty, &a>1\\\text{does not exist}, & a < 0\end{cases}$$
Correct Answer: $B$