Is there a solution to $ \lim_{x\to1^+} \sin\frac{\sqrt{x+1}}{{x^2-1}} $ Is there a solution to the below /Does it exist?
$
\lim_{x\to1^+} \sin\frac{\sqrt{x+1}}{{x^2-1}}
$
When I use conjugate and factor it, I get:
$$
Sin(\sqrt2)
$$
$$
x \neq 0
$$
I found this question asked before and the answer back was it doesn't exist.
Also, found another similar question which does have a solution. So now getting mixed up.
Does the Squeeze Theorem apply to $\lim_{x\to\infty}\sin(\frac{\pi x}{2-3x})$?
$
\sin(\frac{\pi x}{2-3x}) \to \sin(-\frac{\pi}{3}) = -\frac{\sqrt3}{2}
$

As $x \to 1+$, $\sqrt{x+1} \to \sqrt{2}$, but $x^2-1 \to 0$, so $\sqrt{x+1}/(x^2-1) \to \infty$.  $\lim_{t \to \infty} \sin(t)$ does not exist.
