$X_1$, $X_2$, . . . are iid random variables having pdf
$$f_X(x) =\frac{2}{x^3}I_{(1,\infty)}(x)$$
Let
$$V_n=\sqrt{n}\cdot \text{min}{\{X_1, X_2, . . . , X_n}\}$$
Consider the sequence $V_1$, $V_2$, . . . and give the pmf or pdf of the limiting distribution.
I first note that the cdf of $X$ is given by
$$ F_{X}(x)= \begin{cases} 1-\frac{1}{x^2} & x \gt 1 \\ 0 & x\leq 1 \\ \end{cases} $$
We have
$$\begin{align*} F_{V_n}(v) &=\mathsf P(V_n\leq v)\\\\ &=\mathsf P(\sqrt{n}\cdot \text{min}{\{X_1, X_2, . . . , X_n}\}\leq v)\\\\ &=\mathsf P\left(\text{min}{\{X_1, X_2, . . . , X_n}\}\leq \frac{v}{\sqrt{n}}\right)\\\\ &=1-\mathsf P\left(X_1\gt \frac{v}{\sqrt{n}}\right)^n\\\\ &=1-\left(1-\mathsf P\left(X_1 \leq \frac{v}{\sqrt{n}}\right)\right)^n\\\\ &=1-\left(1-\left(1-\frac{1}{\left(\frac{v}{\sqrt{n}}\right)^2}\right)\right)^n\\\\ &=1-\left(\frac{n}{v^2}\right)^n \end{align*}$$
Altogether, we have
$$ F_{V_n}(t)= \begin{cases} 1-\left(\frac{n}{v^2}\right)^n & v\gt \sqrt{n} \\ 0 & v\leq \sqrt{n} \\ \end{cases} $$
and so for all $v\in\mathbb{R}$
$$\lim_{n\rightarrow\infty} F_{V_n}(t)=1$$
Since there is not a valid cdf equal to $1$ except at points of discontinuity, a limiting distribution does not exist.
Is this a valid solution?