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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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    $\begingroup$ Looks good! No mistakes I can see. $\endgroup$ – user837206 Nov 2 '20 at 3:04
  • $\begingroup$ It would be a lot easier if you proved an equivalent case using $b=\sqrt{a}$ and $y=\sqrt{x}$. $\endgroup$ – herb steinberg Nov 2 '20 at 4:52
  • $\begingroup$ @herbsteinberg: on second thoughts, "a lot easier" is not so true (it just spares one or two lines), and this approach requires the knowledge of - continuity of the square root and - a theorem on continuity of composed functions. $\endgroup$ – Yves Daoust Nov 4 '20 at 10:37
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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 $\left| \sqrt{x} + \sqrt{a} \right|$?

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