Limit of sin(sqrt(x+1)) − sin(sqrt(x))
Evaluate the limit: $$ L = \lim_{x \to +\infty} \left(\sin\sqrt{x+1} - \sin\sqrt x\right). $$
Explanation of Method
Since $\sqrt{x+1}$ and $\sqrt x$ both grow without bound, neither $\sin\sqrt{x+1}$ nor $\sin\sqrt x$ individually has a limit — each keeps oscillating. The sum-to-product identity rewrites their difference as a product of a bounded cosine factor and a sine factor whose argument, $\frac{\sqrt{x+1}-\sqrt x}{2}$, shrinks to $0$ as $x\to\infty$. Once the second factor is shown to vanish, the Squeeze Theorem finishes the argument regardless of how the bounded cosine factor behaves.
- Apply the sum-to-product identity $\sin a - \sin b = 2\cos\left(\frac{a+b}2\right)\sin\left(\frac{a-b}2\right)$.
- Show that $\sqrt{x+1}-\sqrt x \to 0$ as $x\to\infty$, by rationalizing.
- Bound the whole expression using $|\cos(\cdot)|\le1$ and apply the Squeeze Theorem.
Step 1 · Apply the sum-to-product identity
With $a=\sqrt{x+1}$ and $b=\sqrt x$:
$$ \sin\sqrt{x+1}-\sin\sqrt x = 2\cos\left(\frac{\sqrt{x+1}+\sqrt x}{2}\right)\sin\left(\frac{\sqrt{x+1}-\sqrt x}{2}\right). $$
Step 2 · Show the difference $\sqrt{x+1}-\sqrt x$ vanishes
Rationalize by multiplying and dividing by the conjugate:
$$ \sqrt{x+1}-\sqrt x = \frac{(x+1)-x}{\sqrt{x+1}+\sqrt x} = \frac{1}{\sqrt{x+1}+\sqrt x}. $$
As $x\to+\infty$, the denominator $\sqrt{x+1}+\sqrt x \to +\infty$, so:
$$ \sqrt{x+1}-\sqrt x = \frac{1}{\sqrt{x+1}+\sqrt x} \longrightarrow 0. $$
Consequently, the argument of the sine factor in Step 1 also tends to $0$:
$$ \frac{\sqrt{x+1}-\sqrt x}{2} \longrightarrow 0 \qquad \text{as } x \to +\infty. $$
Step 3 · Bound the expression and apply the Squeeze Theorem
Since $|\cos(\cdot)|\le1$ for every real argument, no matter how the argument $\frac{\sqrt{x+1}+\sqrt x}{2}$ itself behaves (it grows without bound and $\cos$ of it keeps oscillating), we always have:
$$ \left|\sin\sqrt{x+1}-\sin\sqrt x\right| = 2\left|\cos\left(\frac{\sqrt{x+1}+\sqrt x}{2}\right)\right|\cdot\left|\sin\left(\frac{\sqrt{x+1}-\sqrt x}{2}\right)\right| \le 2\left|\sin\left(\frac{\sqrt{x+1}-\sqrt x}{2}\right)\right|. $$
Since the argument of this remaining sine tends to $0$ (Step 2) and $\sin(u)\to0$ as $u\to0$, the right-hand side tends to $0$. By the Squeeze Theorem:
$$ L = \lim_{x\to+\infty}\left(\sin\sqrt{x+1}-\sin\sqrt x\right) = 0. $$
Summary Table
| Quantity | Behavior as $x \to +\infty$ |
|---|---|
| $\sqrt{x+1}-\sqrt x$ | $=\dfrac{1}{\sqrt{x+1}+\sqrt x}\to0$ |
| $\cos\left(\dfrac{\sqrt{x+1}+\sqrt x}{2}\right)$ | bounded, oscillates in $[-1,1]$ |
| $\sin\sqrt{x+1}-\sin\sqrt x$ | $\to0$ (squeezed) |
Additional Notes
- The Squeeze Theorem is essential here because the cosine factor $\cos\left(\frac{\sqrt{x+1}+\sqrt x}{2}\right)$ never itself settles down to a single value as $x\to\infty$ — it keeps oscillating forever; the argument only works because that oscillating factor is multiplied by something that vanishes, regardless of the oscillation's behavior.
- The key algebraic fact making this work is that $\sqrt{x+1}-\sqrt x\to0$ even though both $\sqrt{x+1}$ and $\sqrt x$ individually diverge to $+\infty$ — their difference shrinks because the "gap" between consecutive square roots narrows as the numbers being rooted grow.
- This is a good template for limits of the form $\lim_{x\to\infty}\left(\sin(f(x))-\sin(g(x))\right)$ whenever $f(x)-g(x)\to0$: the sum-to-product identity plus the Squeeze Theorem will always show the limit is $0$, regardless of how wildly $f$ and $g$ individually behave.
Result
The value of the limit is: