I need to calculate limit: $\lim_{x \rightarrow \infty} (\sin {\sqrt{1+x}} - \sin {\sqrt{x}})$
I was thinking of using the formula for $\sin \alpha - \sin \beta$, but what can be the next step?
2026-04-08 06:27:57.1775629677
Limit $\lim_{x \rightarrow \infty} (\sin {\sqrt{1+x}} - \sin {\sqrt{x}})$
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$$ \sin\sqrt{x+1}-\sin\sqrt{x}= 2\cos\frac{\sqrt{x+1}+\sqrt{x}}{2} \sin\frac{\sqrt{x+1}-\sqrt{x}}{2} $$ Now $$ \lim_{x\to\infty}\frac{\sqrt{x+1}-\sqrt{x}}{2}=0 $$ and $$ \cos\frac{\sqrt{x+1}+\sqrt{x}}{2} $$ is bounded.