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

[Amad27]

Hello:

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

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,

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

 

[Euge]

Hello:

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

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,

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

Hi Olok,

I'm not sure what you're doing in you're analysis but here's an outline for proving the result. Observe that

\(\displaystyle \left|\frac{x}{1 + \sin^2 x}\right| \le |x| \quad \text{for all $x$}.\)

Hence, given $\varepsilon > 0$, setting $\delta = \varepsilon$ forces

\(\displaystyle \left|\frac{x}{1 + \sin^2 x} - 0\right| < \varepsilon\)

for all $x$ such that $0 < |x| < \delta$.

Fill in the details of this argument.

 

[Amad27]

Hi Olok,

I'm not sure what you're doing in you're analysis but here's an outline for proving the result. Observe that

\(\displaystyle \left|\frac{x}{1 + \sin^2 x}\right| \le |x| \quad \text{for all $x$}.\)

Hence, given $\varepsilon > 0$, setting $\delta = \varepsilon$ forces

\(\displaystyle \left|\frac{x}{1 + \sin^2 x} - 0\right| < \varepsilon\)

for all $x$ such that $0 < |x| < \delta$.

Fill in the details of this argument.

Yes I did recognize that.

I know you can conclude if $|x| = \delta = \epsilon$ it will conclude the proof, but I was thinking if my way could work.

Can you tell me, which part is confusing? I really want to try it the way I was doing it.

Thanks!

 

[Euge]

Yes I did recognize that.

I know you can conclude if $|x| = \delta = \epsilon$ it will conclude the proof, but I was thinking if my way could work.

Can you tell me, which part is confusing? I really want to try it the way I was doing it.

Thanks!

One of the main issues is the inequality

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

Since the sine is odd, the fractions on the left- and right-hand sides of the inequality are equal. So the above cannot hold.

 

[blue_raver22]

Therefore, by the definition of a limit, we can conclude that $\displaystyle \lim_{x \to 0}\frac{x}{1 + \sin^2(x)} = 0$.