Prove that, if $x_n \geq 0$, $a > 0$, $n\in \mathbb{N}$, $\lim_{x\to\infty} x_n = a$ then $\lim_{x\to\infty} \sqrt{x_n} = \sqrt{a}$

Question

Prove that, if $x_n \geq 0$, $a > 0$, $n\in \mathbb{N}$, $\lim_{n\to\infty} x_n = a$ then

$$\lim_{n\to\infty} \sqrt{x_n} = \sqrt{a}$$

It looks kind of obvious but I'm having trying to prove it.

My attempt:

Let assume the opposite That is, if $x_n \geq 0$, $a > 0$, $n\in \mathbb{N}$, $\lim_{n\to\infty} x_n = a$ then

$$\lim_{n\to\infty} \sqrt{x_n} = b \neq \sqrt{a}$$

Then it follows that $b^2 \neq a$. But, by applying one of the properties of arithmetic with limits we have that

$$a = \lim_{n\to\infty} x_n = \lim_{n\to\infty} \sqrt{x_n}\cdot \lim_{n\to\infty} \sqrt{x_n} = b \cdot b = b^2 $$

which contradicts our first assumption. Then it follows that:

$$\lim_{n\to\infty} \sqrt{x_n} = b \neq \sqrt{a}$$

Is this okay? I'm not quite convinced.

Answer 1

Fix $\varepsilon > 0$. Since $\lim_{n \to \infty} x_n = a > 0$, there exists $N \in \mathbb{N}$ such that $$\left| x_n - a\right| < \sqrt{a}\varepsilon$$ for each $n > N$. Hence for $n > N$ $$\left| \sqrt{x_n} - \sqrt{a}\right| = \frac{1}{\left| \sqrt{x_n} + \sqrt{a}\right|}\left| x_n - a\right| < \frac{1}{\sqrt{a}}\sqrt{a}\varepsilon = \varepsilon$$ Since $\varepsilon$ was choosen arbitrarily, we conclude $$\lim_{n \to \infty} \sqrt{x_n} = \sqrt{a}$$

Answer 2

Your proof is not correct because you are assuming that $\lim_{n\to\infty}\sqrt{x_n} = b$ exists.

But you can prove it using that $t\to t^2$ is strictly increasing on $(0, \infty)$. If $0 < \varepsilon < \sqrt a$ then $$ 0 < \sqrt a - \varepsilon < \sqrt a < \sqrt a + \varepsilon \\ \Longrightarrow 0 < (\sqrt a - \varepsilon)^2 < a < (\sqrt a + \varepsilon)^2 \, . $$ From $\lim_{n\to\infty} x_n = a$ it follows that there is a $N \in \Bbb N$ such that $$ 0 < (\sqrt a - \varepsilon)^2 < x_n < (\sqrt a + \varepsilon)^2 \quad \text{ for } n \ge N $$ which implies that $$ 0 < \sqrt a - \varepsilon < \sqrt{x_n} < \sqrt a + \varepsilon \quad \text{ for } n \ge N \, . $$

So this is a special case of the fact that the inverse function of a strictly increasing function on an interval is continuous, see for example Continuity of an inverse function..

Answer 3

It is sufficient notice that square root is a continuous function on $[0,+\infty) $.