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,
$\delta = \min(1,\epsilon \cdot (\sin^2(-1) + 1)) \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space  \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \blacksquare$
 A: Remarks on your new approach
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
$$ 
|x|\le |x|\left|\sin^2(x)+1\right|
$$
$$ 
\frac{|x|}{\left|\sin^2(x)+1\right|}\le |x|
$$
Method 1
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 
$$ \lim_{x\to 0} \frac{x}{\sin^2(x)+1} =0 $$
A: 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 $|x|<\delta = \epsilon$?
