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Can the limit $$\lim_{x\to1^+} \sin\frac{\sqrt{x+1}}{{x^2-1}}$$ be evaluated? My attempt employed the squeeze theorem and obtained $$\frac{\sqrt{(-1+1)}}{x^2-1} \lt \sin\frac{\sqrt{x+1}}{{x^2-1}}\lt \frac{\sqrt{(1+1)}}{x^2-1} $$ and from there conclued that $$\lim_{x\to1^+} \sin\frac{\sqrt{x+1}}{{x^2-1}} = \infty$$

But I've never seen it used like this (probably because it can't be...) Any thoughts?

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    The limit does not exist. – 2017-01-27
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    The sine function is bounded by $1$; so, the limit, if it existed, would need to be likewise. That said, the limit does not exist, but rather fluctuates between $-1$ and $1$ as $x\to 1^+$ – 2017-01-27

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