Let $\{X_n\}_{n\geq 1}$ be a sequence of random variables s.t $X_n \to a$ in probability, where $a>0$ is a fixed constant. I'm trying to show the following: $$\sqrt{X_n} \to \sqrt{a}$$ and $$\frac{a}{X_n}\to 1$$ both in probability. I'm here to see if my logic was sound. Here's my work
ATTEMPT
For the first part, we have $$|\sqrt{X_n}-\sqrt{a}|<\epsilon \impliedby |X_n-a|<\epsilon|\sqrt{X_n}+\sqrt{a}|=\epsilon|(\sqrt{X_n}-sqrt{a})+2\sqrt{a}|$$ $$\leq \epsilon|\sqrt{X_n}-\sqrt{a}|+2\epsilon\sqrt{a}<\epsilon^2+2\epsilon\sqrt{a}$$ Notice that $$\epsilon^2+2\epsilon\sqrt{a}>\epsilon\sqrt{a}$$ It follows then that $$P(|\sqrt{X_n}-\sqrt{a}|\leq \epsilon)\geq P(|X_n-a|\leq \epsilon\sqrt{a})\to 1 \;\;as\;n\to\infty$$ $$\implies \sqrt{X_n}\to\sqrt{a} \;\;in\;probability$$
For the second part, we have $$|\frac{a}{X_n}-1|=|\frac{X_n-a}{X_n}|<\epsilon \impliedby |X_n-a|<\epsilon|X_n|$$ Now, since $X_n \to a$ as $n \to \infty$, we have that $X_n$ is a bounded sequence. In other words, there exists a real number $M<\infty$ s.t $|X_n|\leq M$. Thus, $$|X_n-a|<\epsilon|X_n|\impliedby |X_n-a|<\epsilon M$$ Looking at it in probability, we have $$P(|\frac{a}{X_n}-1|>\epsilon)=P(|X_n-a|>\epsilon|X_n|)\leq P(|X_n-a|>\epsilon M)\to 0 \;\;as\;n\to\infty$$
I'm pretty confident in the first one, but am pretty iffy on the second. Was my logic sound?