Is this an accurate limit proof for sine?

Question

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

$$\left|\frac x{1 + \sin^2(x)}\right| < \epsilon$$

$$|x| < \delta$$

Let's require $|x| < 1$ so therefore,

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

Therefore we get,

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

We must require,

$$\epsilon = \frac\delta{\sin^2(1) + 1}$$

Therefore, $\delta = \epsilon (\sin^2(1) + 1)$ BUT WE MUST CONSIDER $|x| < 1$ so finally,

$$\delta = \min(1, \epsilon(\sin^2(1) + 1))$$

Correct?

thanks!

Answer 1

You appear to be asserting that

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

implies that

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

when in fact the opposite is true. But it is certainly true that $\sin^2(|x|) + 1 \geq 1.$ This enables you to write a very useful inequality comparing $\left|\frac{x}{1 + \sin^2 x}\right|$ and $|x|$.