How to prove $\underset{n\to \mathrm{\infty}}{lim\u2006sup}|\mathrm{sin}(n)|=1$?

Ayaan Barr
2022-07-04
Answered

How to prove $\underset{n\to \mathrm{\infty}}{lim\u2006sup}|\mathrm{sin}(n)|=1$?

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poquetahr

Answered 2022-07-05
Author has **18** answers

See this article, a special case is that there is two increasing sequences of odd positive integers $({p}_{n}),({q}_{n})$ such that

$|\frac{\pi}{2}-\frac{{p}_{n}}{{q}_{n}}|\le \frac{1}{{q}_{n}^{2}},\phantom{\rule{1em}{0ex}}n>1.$

Note that $\mathrm{sin}|x|=|\mathrm{sin}x|$ for $x\in [0,\pi ]$, then

$\mathrm{sin}|\frac{{q}_{n}\pi}{2}-{p}_{n}|=|\mathrm{cos}{p}_{n}|<\frac{1}{{q}_{n}}\to 0.$

therefore $|\mathrm{sin}{p}_{n}|\to 1$

$|\frac{\pi}{2}-\frac{{p}_{n}}{{q}_{n}}|\le \frac{1}{{q}_{n}^{2}},\phantom{\rule{1em}{0ex}}n>1.$

Note that $\mathrm{sin}|x|=|\mathrm{sin}x|$ for $x\in [0,\pi ]$, then

$\mathrm{sin}|\frac{{q}_{n}\pi}{2}-{p}_{n}|=|\mathrm{cos}{p}_{n}|<\frac{1}{{q}_{n}}\to 0.$

therefore $|\mathrm{sin}{p}_{n}|\to 1$

asked 2021-10-13

Given that $\underset{x\to a}{lim}f\left(x\right)=0,\underset{x\to a}{lim}g\left(x\right)=0,\underset{x\to a}{lim}h\left(x\right)=1,\underset{x\to a}{lim}p\left(x\right)=\mathrm{\infty},\underset{x\to a}{lim}q\left(x\right)=\mathrm{\infty}$ which of the following limits are indeterminate forms? Fro those that are not an indeterminate form, evaluate the limit possible. $\underset{x\to a}{lim}[p\left(x\right)-q\left(x\right)]$

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Evaluate the limits, using algebra and/or limit properties as needed. $\underset{x\to 2}{lim}\frac{3{x}^{2}-5}{{x}^{2}+3x-3}$

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How to solve.

$lim(1-1/{h}^{2})/(1-1/h)$

$h\to 0$

$lim(1-1/{h}^{2})/(1-1/h)$

$h\to 0$

asked 2020-10-27

Use the conjugate process
$\underset{x\to 0}{lim}\frac{\sqrt{{x}^{2}+5}-\sqrt{5-{x}^{2}}}{x}.$

asked 2022-06-25

Where is the mistake in this solution of $\underset{x\to 1}{lim}\frac{1-{x}^{2}}{\mathrm{sin}(\pi x)}$?

The answer ought to be $\frac{2}{\pi}$, but I end up with 0:

$\underset{x\to 1}{lim}\frac{1-{x}^{2}}{\mathrm{sin}(\pi x)}=$ $\underset{y\to 0}{lim}\frac{1-(y+1{)}^{2}}{\mathrm{sin}(\pi (y+1))}=$ $\underset{y\to 0}{lim}\frac{\pi (y+1)}{\mathrm{sin}(\pi (y+1))}\frac{1-(y+1{)}^{2}}{\pi (y+1)}=$ $\underset{y\to 0}{lim}\frac{1-(y+1{)}^{2}}{\pi (y+1)}=0$

Where and why is my solution incorrect?

The answer ought to be $\frac{2}{\pi}$, but I end up with 0:

$\underset{x\to 1}{lim}\frac{1-{x}^{2}}{\mathrm{sin}(\pi x)}=$ $\underset{y\to 0}{lim}\frac{1-(y+1{)}^{2}}{\mathrm{sin}(\pi (y+1))}=$ $\underset{y\to 0}{lim}\frac{\pi (y+1)}{\mathrm{sin}(\pi (y+1))}\frac{1-(y+1{)}^{2}}{\pi (y+1)}=$ $\underset{y\to 0}{lim}\frac{1-(y+1{)}^{2}}{\pi (y+1)}=0$

Where and why is my solution incorrect?