سوال ۳۲ کنکور ارشد مهندسی کامپیوتر ۱۴۰۱

Apply the sum-to-product identity $\sin A-\sin B=2\cos\!\left(\frac{A+B}{2}\right)\sin\!\left(\frac{A-B}{2}\right)$ with $A=\sqrt{x+1}$ and $B=\sqrt{x}$: $\sin\sqrt{x+1}-\sin\sqrt{x}=2\cos\!\left(\frac{\sqrt{x+1}+\sqrt{x}}{2}\right)\sin\!\left(\frac{\sqrt{x+1}-\sqrt{x}}{2}\right).$

The gap between the square roots vanishes (rationalize): $\sqrt{x+1}-\sqrt{x}=\frac{(x+1)-x}{\sqrt{x+1}+\sqrt{x}}=\frac{1}{\sqrt{x+1}+\sqrt{x}}\xrightarrow[x\to\infty]{}0.$

Hence the sine factor tends to $\sin(0)=0$, while the cosine factor is bounded: $\left|2\cos\!\left(\frac{\sqrt{x+1}+\sqrt{x}}{2}\right)\right|\le 2.$

A bounded quantity times one that goes to $0$ goes to $0$ (squeeze): $\left|\sin\sqrt{x+1}-\sin\sqrt{x}\right|\le 2\left|\sin\!\left(\frac{1}{2(\sqrt{x+1}+\sqrt{x})}\right)\right|\to 0.$

Therefore the limit equals $0$, matching option 1.

Exam trick: Although $\cos(\cdot)$ oscillates without limit, multiplying a bounded oscillating factor by something that $\to 0$ kills it. Don't be tempted by the "limit does not exist" choice — the oscillation is dampened to zero.