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

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 $|x|<\delta = \epsilon$?
