$\textbf{The Problem:}$ Suppose that $X_1,X_2,\dots$ are iid random variables with PDF $$f(x)=\begin{cases}x^{-2}&\text{if }x\geq1\\0&\text{otherwise.} \end{cases}$$ Let $M_n=\max\{X_1,\dots,X_n\}$. Show that $M_n/n$ converges in distribution, and identify the CDF of the limiting distribution.
$\textbf{My Thoughts:}$ The independence of the random variables implies that $$\begin{align*}\mathsf P(M_n\leq nx)&=\mathsf P(X_1\leq nx)^n\\&=\left(\int_{1}^{nx}\frac{1}{y^2}dy\right)^{n}\\&=\left(1-\frac{1}{nx}\right)^{n} \end{align*}$$ for all $x$ such that $nx>1$. The above converges to $\large e^{-x^{-1}}$ as $n\to\infty$. Therefore I conclude that the CDF of the limiting distribution is given by $$\mathsf F(x)=\begin{cases}e^{-x^{-1}}&\text{if }x>0\\0&\text{otherwise.} \end{cases}$$ Therere the PDf is given by $$\mathsf f_1(x)=\begin{cases}\displaystyle\frac{e^{-x^{-1}}}{x^2}&\text{if }x>0\\0&\text{otherwise.} \end{cases}$$ Then I compute that $$\int_{-\infty}^{\infty}\mathsf f_1(x)dx=1.$$ The proposed CDF gives an appropriate PDF, however I think I made a mistake in taking $x>0$ and hence in the overall calculation.
Is my reasoning above correct?
Thank you for your time and any feedback provided is much appreciated.
Your computation is fine. If you feel uncomfortable for introducing the condition $x > 0$, it may help to utilize the indicator notation $\mathbf{1}(\cdots)$, which is defined as
$$ \mathbf{1}(\cdots) = \begin{cases} 1, &\text{if $\cdots$ holds}; \\ 0, &\text{if $\cdots$ does not hold}. \end{cases} $$
Using this, we have $f(x) = x^{-2}\mathbf{1}(x \geq 1)$, and so,
$$ \mathsf{P}(X \leq x) = \int_{-\infty}^{x} f(t) \, \mathrm{d}t = \int_{-\infty}^{x} t^{-2}\mathbf{1}(t \geq 1) \, \mathrm{d}t = (1 - x^{-1}) \mathbf{1}(x \geq 1). $$
Then it follows that
$$ \mathsf{P}(M_n/n \leq x) = \mathsf{P}(X_1 \leq nx)^n = \left(1 - \frac{1}{nx}\right)^n \mathbf{1}(x \geq 1/n) \xrightarrow[n\to\infty]{} e^{-1/x} \mathbf{1}(x > 0). $$
This naturally produces the range of $x$ for which the limiting CDF coincides $e^{-1/x}$.