$X_1$, $X_2$, . . . are iid random variables having pdf $$f_X(x) =\frac{2}{x^3}I_{(1,\infty)}(x)$$ Let $$T_n= \frac{\text{max}{\{X_1, . . . , X_n}\}}{\sqrt{n}}$$ (a) Consider the sequence $T_1$, $T_2$, . . . 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
$$\begin{align*} F_{T_n}(t) &=\mathsf P(T_n\leq t)\\\\ &=\mathsf P\left(\frac{\text{max}{\{X_1,...,X_n}\}}{\sqrt{n}}\leq t\right)\\\\ &=\mathsf P\left(\text{max}{\{X_1,...,X_n}\}\leq \sqrt{n}\cdot t\right)\\\\ &=\mathsf P\left(X_1\leq \sqrt{n}\cdot t,...,X_n\leq\sqrt{n}\cdot t\right)\\\\ &=\prod_{i=1}^n \mathsf P\left(X_i \leq \sqrt{n}\cdot t\right)\\\\ &=\left(1-\frac{1}{(\sqrt{n}\cdot t)^2}\right)^n \end{align*}$$
Altogether, we have
$$ F_{T_n}(t)= \begin{cases} \left(1-\frac{1}{(\sqrt{n}\cdot t)^2}\right)^n & t \gt \frac{1}{\sqrt{n}} \\ 0 & x\leq \frac{1}{\sqrt{n}} \\ \end{cases} $$
Using the property that
$$\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^n\rightarrow e^x$$
we get
$$ \lim_{n\rightarrow\infty}F_{T_n}(t)= \begin{cases} e^{-\frac{1}{t^2}} & t \gt 0 \\ 0 & t \lt 0 \\ \end{cases} $$
Then
$$\frac{d}{dt}\left(e^{-\frac{1}{t^2}}\right)=\frac{2e^{-\frac{1}{t^2}}}{t^3}$$
Finally, we get
$$f_T(t)=\frac{2e^{-\frac{1}{t^2}}}{t^3}I_{(0,\infty)}(t)$$
Is this a valid solution?