Could someone help to check my proof, I'm not sure if they are rigorous:
Given ${a_n}$, ${b_n}$, and ${c_n}$ be sequences of real numbers and let k be a constant, $\lim_{n \to \infty} a_n = \infty$, $\lim_{n \to \infty} b_n = \infty$, and $\lim_{n \to \infty} c_n = -\infty$.
To prove $\lim_{n \to \infty} a_nb_n = \infty$, we fix any $M \in \mathbb{R} $. Since $\lim_{n \to \infty} a_n = \infty$, $\exists N_1 \in \mathbb{N}$ such that $a_n \ge \sqrt{M}$$\forall n \ge N_1$. Similarly, $\exists N_2 \in \mathbb{N}$ such that $b_n \ge \sqrt{M}$$\forall n \ge N_2$. Let $N=max${$N_1,N_2$}. Then, $a_nb_n \ge M$$\forall n \ge N$, which implies the claim.
To prove $\lim_{n \to \infty} a_nc_n = -\infty$, we fix any $M \in \mathbb{R} $. Since $\lim_{n \to \infty} a_n = \infty$, $\exists N_1 \in \mathbb{N}$ such that $a_n \ge \sqrt{M}$$\forall n \ge N_1$. Similarly, $\exists N_2 \in \mathbb{N}$ such that $c_n \le \sqrt{M}$$\forall n \ge N_2$. We can also write it as $-c_n \ge \sqrt{M}$. Let $N=max${$N_1,N_2$}. Then, $-a_nc_n \ge -M$ which is $a_nc_n \le M$$\forall n \ge N$, which implies the claim.
To prove $\lim_{n \to \infty} ka_n = \infty$ if $k \gt 0$, and $\lim_{n \to \infty} ka_n = -\infty$ if $k \lt 0$, we fix any $M \in \mathbb{R} $. Since $\lim_{n \to \infty} a_n = \infty$, $\exists N \in \mathbb{N}$ such that $a_n \ge \frac{M}{k}$$\forall n \ge N$. If $k \gt 0$, then $ka_n \ge M$, else if $k \lt 0$, then $ka_n \le M$, which implies the claim.