Prove using the $\epsilon-\delta$ definition that $\lim_{x \to a} \sin \sqrt{x} = \sin \sqrt{a}$ for any $a > 0$, $a \in \mathbb{R}$.

Question

Scratch Work

Suppose $\lim_{x \to a} \sin \sqrt{x} = \sin \sqrt{a}$. Then for every $\epsilon > 0$, we aim to find a $\delta > 0$ such that

$$ 0 < |x-a| < \delta \Rightarrow \left|\sin \sqrt{x} - \sin \sqrt{a}\right| < \epsilon $$

Note that

\begin{align*} \sin x &\le x \\ \cos x &\le 1 \end{align*}

for every $x \ge 0$.

Also

\begin{align*} |x-a| &= |\sqrt x - \sqrt a| \cdot |\sqrt x + \sqrt a| \end{align*}

Proof

Choose $\delta = \min\{a, \epsilon \sqrt {a}\}$. We have

\begin{align*} 0 < |x-a| < \delta &\Rightarrow \left|\sin \sqrt{x} - \sin \sqrt{a}\right| \\ &= \left|2\cos\left(\frac {\sqrt{x} + \sqrt{a}} 2\right) \cdot \sin\left(\frac {\sqrt{x} - \sqrt{a}} 2 \right) \right| \\ &\le 2 \cdot 1 \cdot \frac {\left|\sqrt {x} - \sqrt {a} \right|} 2 \\ &= \frac {|x-a|} {|\sqrt x + \sqrt a|} \\ &\le \frac 1 {\sqrt a} \cdot \epsilon \sqrt a \\ &= \epsilon \end{align*}

I am more concerned with my workings which led me to the proof. Is there any mistake in my manipulations? Any advice would be greatly appreciated!

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Answer 1

I agree with user837206’s comment: your post is correct.

I suggest some tweaks:

You start with “Suppose $\lim_{x \to a} \sin \sqrt{x} = \sin \sqrt{a}$.” That’s the statement you are trying to prove. It’s just some notes, not the actual proof, so it’s not technically wrong. But make sure you are clear about what you have already proved and what you still need to prove.

You follow with “for every $\epsilon > 0$, we aim to find a $\delta > 0$ …”. I would say we can find such a $\delta > 0$.

Turning to the actual proof, you can strengthen most of your inequalities:

For every $x > 0$, $\sin x < x$, so

$$\left| 2 \cos\left(\frac{\sqrt{x} + \sqrt{a}}{2}\right) \sin\left(\frac{\sqrt{x} - \sqrt{a}}{2}\right) \right| < 2 \cdot 1 \cdot \frac{\left|\sqrt{x} - \sqrt{a}\right|}{2}.$$

By using either of these inequalities:

\begin{align*} |x - a| < \delta \leq a &\implies x > 0 \\ &\implies \frac{1}{\left| \sqrt{x} + \sqrt{a} \right|} < \frac{1}{\sqrt{a}} \tag{1} \\ |x - a| < \delta \leq \epsilon\sqrt{a} &\implies |x - a| < \epsilon\sqrt{a} \tag{2} \end{align*}

We can make this inequality strict:

$$\frac{|x - a|}{\left| \sqrt{x} + \sqrt{a} \right|} < \frac{1}{\sqrt{a}} \cdot \epsilon\sqrt{a}.$$

Actually, do we really need the absolute value in $\left| \sqrt{x} + \sqrt{a} \right|$?