Is there a solution to $ \lim_{x\to1^+} \sin\frac{\sqrt{x+1}}{{x^2-1}} $

Question

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} $

Answer 1

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.

Answer 2

I think the difference is that $ \frac{\pi x}{2-3x}$ converges as $x \rightarrow \infty$ (as in the other question) but $\frac{\sqrt{x+1}}{x^2-1}$ doesn't as $x \rightarrow 1^+$. In fact $\frac{\sqrt{x+1}}{x^2-1} \rightarrow \infty$ as $x \rightarrow 1^+$ so hopefully you can see that $\sin(\frac{\sqrt{x+1}}{x^2-1})$ won't converge as it will just keep cycling round.