Let $x_n \geq 0$ for all $n \in \mathbf{N}$. Show that if $(x_n) \longrightarrow x$, Then $(\sqrt{x_n}) \longrightarrow \sqrt{x}$.

For the definitions of sequences and what it means to for a sequence to converge, see this, for subsequences see this.
For the “show the limit” template and an example, see this.

Problem Discussion

We’ll follow a similar approach to proving the example in here. We want to show that $\lim (\sqrt{x_n}) = x$. So we want to find $N$ such that for any $\epsilon > 0$,

$$ \begin{align*} \lvert \sqrt{x_n} - \sqrt{x} \rvert < \epsilon. \end{align*} $$

The trick here is to multiply the left hand side by the conjugate!!!

The numerator is exactly what we want. The denominator however might be 0 since $x$ is non-negative. Therefore, we’ll just assume that $x > 0$ for now to continue the proof and we’ll handle the case when $x = 0$ separately. Also since the denominator is positive then we can just write:

$$ \begin{align*} \lvert \sqrt{x_n} - \sqrt{x} \rvert &= \frac{\lvert x_n - x \rvert}{\sqrt{x_n} + \sqrt{x}}. \end{align*} $$

So far so good but we also know that the sequence $(x_n)$ converges to $x$ and therefore, it is bounded. This means that there exists some bound $M$ such that $|(x_n)| \leq M$ and so $\sqrt{|(x_n)|} \leq \sqrt{M}$ (why?). Since $x_n \geq 0$, then $M \geq 0$ and we can remove the term to the get the following inequality:

$$ \begin{align*} \lvert \sqrt{x_n} - \sqrt{x} \rvert &= \frac{\lvert x_n - x \rvert}{\sqrt{x_n} + \sqrt{x}} \\ &< \frac{\lvert x_n - x \rvert}{\sqrt{x}} \\ \end{align*} $$

We can set $\lvert x_n - x \rvert$ to anything we like. Particularly we want to set it to a value such that the terms would cancel out and we would get $\epsilon$ at the end. So let it be $\epsilon \sqrt{x}$. Remember that $x$ is a constant so that’s okay.

$$ \begin{align*} \lvert \sqrt{x_n} - \sqrt{x} \rvert &< \frac{\lvert x_n - x \rvert}{\sqrt{x}} \\ &< \frac{\sqrt{x}\epsilon}{\sqrt{x}} = \epsilon. \end{align*} $$

Formal Proof

To show that $\lim(\sqrt{x_n}) = \sqrt{x}$, let $\epsilon > 0$ be arbitrary. We want to prove that there exists some $n \geq N$ such that,

$$ \begin{align*} \lvert \sqrt{x_n} - \sqrt{x} \rvert < \epsilon \end{align*} $$

To do so, we’ll consider two cases for when $x = 0$ and $x > 0$. For case 1, suppose $x > 0$. We are given that $\lim x_n = x$. This means that there exists some $n \geq N$ such that

$$ \begin{align*} \lvert x_n - x \rvert < \epsilon \sqrt{x} \quad \text{whenever $n \geq N$}. \end{align*} $$

Now assume that $n \geq N$, then

$$ \begin{align*} &\lvert \sqrt{x_n} - \sqrt{x} \rvert * \frac{\lvert \sqrt{x_n} + \sqrt{x} \rvert}{\lvert \sqrt{x_n} + \sqrt{x} \rvert} \\ &= \frac{\lvert x_n - x \rvert}{\lvert \sqrt{x_n} + \sqrt{x} \rvert} \\ &= \frac{\lvert x_n - x \rvert}{\sqrt{x_n} + \sqrt{x}} \\ &< \frac{\lvert x_n - x \rvert}{\sqrt{x}} \quad \text{($(x_n)$ is bounded and $x_n \geq 0$)} \\ &< \frac{\lvert x_n - x \rvert}{\sqrt{x}} \\ &< \frac{\sqrt{x}{\epsilon}}{\sqrt{x}} = \epsilon, \\ \end{align*} $$

as required. For case 2 when $x = 0$, we want to prove that

$$ \begin{align*} \lvert \sqrt{x_n} - 0 \rvert &< \epsilon \\ \lvert \sqrt{x_n} \rvert &< \epsilon \\ \end{align*} $$

We are given that $\lim x_n = x$ or $\lim x_n = 0$. This means that there exist some $n \geq N$ such that

$$ \begin{align*} \lvert x_n - 0 \rvert = \lvert x_n \rvert &< \epsilon^2. \\ \end{align*} $$

Now assume that $n \geq N$, then

$$ \begin{align*} \lvert \sqrt{x_n} \rvert &= \sqrt{x_n} \quad \text{since $x_n \geq 0$}\\ &= \sqrt{|x_n|} \\ &< \sqrt{\epsilon^2} = \epsilon, \end{align*} $$

as required. $\blacksquare$

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