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It is well kown that $$\lim_{x\to\infty} \sin(x)$$ does not exist (the same for all non-constant periodic function).

Please see Help where it can be deduced (really, or I am wrong?) $$\lim_{x\to\infty} \sin \left(\frac{x^2+5}{x+5}\right)^{1/2} =\lim_{x\to\infty} \sin (x^2-5x+25)^{1/2}$$ How to interpret this?

Community
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Piquito
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  • show that $\sin$ takes the values $\pm 1$ in any neighborhood of $\infty$. – user251257 Jul 15 '15 at 15:57
  • It most certainly does not since $-1\le \sin x\le 1$ for all real $x$. – Tim Raczkowski Jul 15 '15 at 16:00
  • Are you familiar with the definition of a limit? Clearly this cannot exist since the sine function will be oscillating between $-1$ and $1$ as the argument increases. – jameselmore Jul 15 '15 at 16:05
  • Certainly, the only periodic functions that admit such a limit are the constant functions. My main question is really about the comment that follows the question, it is some intriguing to me. – Piquito Jul 15 '15 at 16:09
  • Theorem: If $f$ is a periodic function and $\lim_{x\to \infty}f(x)$ exists then it must be constant.. – Empty Jul 15 '15 at 16:15

3 Answers3

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We have $$ \lim_{n\to \infty} \sin(2\pi n) = 0 $$ and $$ \lim_{n\to \infty} \sin(2\pi n + \frac{\pi}{2}) = 1, $$ so $\lim_{n\to \infty} \sin n$ does not exist. That's why your equality $$ \lim_{x\to\infty} \sin (\frac{x^2+5}{x+5})^{1/2} =\lim_{x\to\infty} \sin (x^2-5x+25)^{1/2} $$ doesn't make sense to me (neither left side nor right side of this equality is defined because limits do not exist).

Virtuoz
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  • Knowing that $a sin (\alpha) - b sin(\beta)\to 0$ make $a sin (\alpha) - b sin(\beta)=h$ then $\frac{a}{b} sin(\alpha) - sin(\beta)= \frac {h}{b}$. But $\frac{a}{b} \to 1$. Then $sin(\alpha) - sin(\beta)\to 0$. My attention is on this last point. – Piquito Jul 15 '15 at 16:23
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If $\sin x$ had some limit $\ell$ as $x \to \infty$, then we would have that for every sequence $(a_n)$ such that $a_n \to \infty$ and hence $$\lim_{n \to \infty} \sin a_n = \ell.$$

However, if $(a_n) = kn\pi$ then the limit is $0$. If the sequence is $(a_n) = \frac{\pi}{2} + kn\pi$ then the limit becomes $1$. Hence the sine function has no limit as $x \to \infty$.

It follows trivially from the fact that $$\lim_{x \to \infty} \sqrt{\frac{x^2+5}{x+5}} = \infty$$ that the limit does not exist.

Zain Patel
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Assume the limit exists$\ =a$. If $a=0$ then we can take $x_n=(2n+1)\cdot \pi/2$ and see that the limit is $0$ not $a$. If $a$ is not zero then we can take $y_n=n\cdot \pi/2$ and see that the limit is $0$. Thus $a$ does not exist.

iadvd
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Adelafif
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