I am not sure how to calculate the limit: $$\lim_{x\to\infty}\cos(\sqrt{x+1})-\cos(\sqrt{x})$$

I applied trigonometric identity to get:

$$L=-\lim_{x\to \infty} 2\sin\left(\dfrac{1}{2}\dfrac{1}{(\sqrt{1+x}-\sqrt{x})}\right) \sin\left(\dfrac{1}{2}\dfrac{1}{(\sqrt{1+x}+\sqrt{x})}\right)$$

Not sure how to proceed using trigonometry.

Also, is there a method using L'Hospital for $\infty-\infty$ form (not for this question obviously!)?


METHODOLOGY $1$: Pre-Calculus Approach

Recalling that $\cos(x)-\cos(y)=-2\sin\left(\frac{x-y}{2}\right)\sin\left(\frac{x+y}{2}\right)$, $|\sin(\theta)|\le |\theta|$, and $|\sin(\theta)|\le 1$, we have

$$\begin{align} \left|\cos(\sqrt{x+1})-\cos(\sqrt{x})\right|&=2\left|\sin\left(\frac{\sqrt{x+1}-\sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x+1}+\sqrt{x}}{2}\right)\right|\\\\ &=2\left|\underbrace{\sin\left(\frac{1}{2\left(\sqrt{x+1}+\sqrt{x}\right)}\right)}_{\text{bounded by its argument in absolute value}}\,\,\,\underbrace{\sin\left(\frac{\sqrt{x+1}+\sqrt{x}}{2}\right)}_{\text{bounded by}\,1 \,\text{in absolute value}}\right|\\\\ &\le \frac{1}{\sqrt{x+1}+\sqrt{x}}\\\\ &\le \frac{1}{2\sqrt{x}}\to 0 \end{align}$$

as $x\to \infty$.

METHODOLOGY $2$: Calculus Approach

Alternatively, we can use the mean value theorem applied to $\cos(\sqrt{x})$. Then, there exists a number $\xi\in (x,x+1)$ such that

$$\cos(\sqrt{x+1})-\cos(\sqrt{x})=-\frac{\sin(\sqrt{\xi})}{2\sqrt{\xi}}\to 0$$

as $x\to \infty$.

  • $\begingroup$ So that would mean limit is zero? $\endgroup$ – samjoe Apr 5 '17 at 5:36
  • $\begingroup$ @samjoe Indeed. $\endgroup$ – Mark Viola Apr 5 '17 at 5:36
  • $\begingroup$ Thank you for answer! Actually I also thought that limit was zero, as second term went zero. But was not able to proceed! $\endgroup$ – samjoe Apr 5 '17 at 5:40
  • $\begingroup$ @samjoe this is because of the "squeeze theorem". $\endgroup$ – Kevin Apr 5 '17 at 6:45
  • $\begingroup$ @samjoe You're welcome. My pleasure. -Mark $\endgroup$ – Mark Viola Apr 5 '17 at 14:11

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