Prove that: $\lim_{x\to 0}\frac{x}{\sin^2(x) + 1} = 0$

Prove

$$\displaystyle \lim_{x\to 0} \frac{x}{\sin^2(x) + 1} = 0$$

The proof:

Let $$|x| \le 1 \implies -1 \le x \le 1$$

$$\displaystyle \frac{|x|}{|\sin^2(x) + 1|} < \epsilon\text{ for }\displaystyle |x| < \delta$$

$$-1 \le x \le 1 \\\implies \sin(-1) \le \sin(x) \le \sin(1) \implies -\sin(1) \le \sin(x) \le \sin(1) \\\implies \sin^2(1) \le \sin^2(x) \le \sin^2(1) \implies |\sin^2(x) + 1| = |\sin^2(1) + 1| \implies \displaystyle |\frac{1}{\sin^2(x) + 1}| = |\frac{1}{\sin^2(1) + 1}|$$

(1) $$|x| < \delta_1$$

(2) $$\displaystyle |\frac{1}{\sin^2(x) + 1}| = |\frac{1}{\sin^2(1) + 1}|$$

(3) $$\displaystyle \frac{|x|}{|\sin^2(x) + 1|} < \frac{\delta_1}{|\sin^2(1) + 1|}$$

(4) $$\displaystyle \frac{|\delta_1|}{|\sin^2(1) + 1|} = \epsilon \implies \delta_1 = (|\sin^2(1) + 1|)(\epsilon)$$

Finally, $\delta = \min(1, (|\sin^2(1) + 1|)(\epsilon)) \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \blacksquare$

Thoughts?

EDIT:

The original proof was indeed terrible, here's a new approach.

Let $|x| < 1 \implies -1 < x < 1$

$\sin^2(-1) + 1 < \sin^2(x) + 1 <\sin^2(1) + 2$

$\implies \displaystyle \frac{1}{\sin^2(-1) + 1} > \frac{1}{\sin^2(x) + 1} > \frac{1}{\sin^2(1) + 1}$

$\implies \displaystyle \frac{1}{\sin^2(-1) + 1} > \frac{1}{\sin^2(x) + 1} \implies \frac{1}{|\sin^2(-1) + 1|} > \frac{1}{|\sin^2(x) + 1|} \implies \frac{1}{|\sin^2(x) + 1|} < \frac{1} {|\sin^2(-1) + 1|}$

$(1) |x| < \delta_1$

$(2) \displaystyle \frac{1}{|\sin^2(x) + 1|} < \frac{1} {|\sin^2(-1) + 1|}$

$(3) \displaystyle \frac{|x|}{|\sin^2(x) + 1|} < \frac{\delta_1} {|\sin^2(-1) + 1|}$

Finally,

$\epsilon(\sin^2(-1) + 1) = \delta_1$

Therefore,

$\delta = \min(1,\epsilon \cdot (\sin^2(-1) + 1)) \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \blacksquare$

• You can't get to $\sin^2(1) \le \sin^2(x)$ from your start. – coffeemath Nov 3 '14 at 18:16
• $a \leq b$ do not implies $a^2 \leq b^2$... take $a = -2, b = 1$ and your inequality $\sin^2 (1)\leq \sin^2(x) \leq \sin^2(1)$ implies $\sin^2(x)=\sin(1)$ which is obviously not always true for $|x| \leq 1$. – Surb Nov 3 '14 at 18:16
• there should probably be an explanation to $-1<x<1 \implies -\sin 1 \le \sin x \le \sin 1$. – mookid Nov 3 '14 at 18:17
• Okay. $-1 < x < 1$ therefore $sin(-1) < sin(x) < sin(1)$ but realize $sin(-1) = -sin(1)$ then square and proceed – Amad27 Nov 3 '14 at 18:28
• You cannot square an inequality when one or both of the sides are negative! $-\sin(1) \leq \sin(x)$ does not imply $\sin^2(1) \leq \sin^2(x)$ (try $x=0$)! The proof has many mistakes and is way to long and complicated. mookids answer below is the way to go here. – Winther Nov 4 '14 at 3:23

This is way too complicated, don't you think?

Why not just say that $$\left| \frac{x}{1+\sin^2 x} \right| \le |x| \le \epsilon$$ as soon as $|x|<\delta = \epsilon$?

• I just want to know if my approach is correct – Amad27 Nov 3 '14 at 18:30
• there seem to be some things not right. just look at the comments under your question. – mookid Nov 3 '14 at 18:38
• What is wrong? I gave a comment to your there as well – Amad27 Nov 3 '14 at 19:24

Since in your attempt, $x\in[-1, 1]$, lets try $x=0$ $$\sin^2(-1)+1\lt \sin^2(0)+1$$ $$\sin^2(-1)\lt \sin^2(0)$$ $$\sin^2(1)\lt 0$$ Is this correct to you? Also if you wish to start from $|x|$, then why not start with the fact that $$|x|\le |x|\left|\sin^2(x)+1\right|$$ $$\frac{|x|}{\left|\sin^2(x)+1\right|}\le |x|$$
Unless your forced to use the definition of the limit, we could simply show that $$\lim_{x\to 0} \frac{x}{\sin^2(x)+1} =\frac{0}{\sin^2(0)+1} = \frac{0}{0+1} = \frac{0}{1} = 0$$ Method 2
If the other answer isn't obvious to you, here's another way to see it. First note that for $x\in\mathbb{R}$ $$0\le\sin^2(x)$$ $$1\le\sin^2(x)+1$$ $$1\le\left|\sin^2(x)+1\right|$$ $$\frac{1}{\left|\sin^2(x)+1\right|}\le 1$$ $$\frac{|x|}{\left|\sin^2(x)+1\right|}\le |x|$$ $$\left|\frac{x}{\sin^2(x)+1}-0\right|\le |x-0|$$ Let $\epsilon \gt 0$ and $\delta=\epsilon$, then $$\left|\frac{x}{\sin^2(x)+1}-0\right|\lt \epsilon \quad \mbox{whenever} \quad 0\lt\left|x-0\right|\lt\delta$$ Therefore $$\left|\frac{x}{\sin^2(x)+1}-0\right|\le |x-0|\lt \delta=\epsilon$$ And by definition $$\lim_{x\to 0} \frac{x}{\sin^2(x)+1} =0$$