About limits of $\sin x$ 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?
 A: 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).
A: 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.
A: 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.
