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How should I determine the following limit?

$\lim _{x\to \infty }\left(\cos\sqrt{x}-\cos\sqrt{x-1}\right)$

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  • $\begingroup$ Hint: Consider all possible ways in which $x$ approaches $\infty$ $\endgroup$ – kishlaya Jan 21 '17 at 11:48
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    $\begingroup$ @kishlaya Sorry but your "hint" is not understandable. $\endgroup$ – Did Jan 21 '17 at 13:10
  • $\begingroup$ @Did Oh! I meant like, consider case wise, $x = k^2, k^2 < x < (k+1)^2,...$. But nevertheless, I will try to come up with better explanations next time. My apologies. $\endgroup$ – kishlaya Jan 22 '17 at 7:58
  • $\begingroup$ @kishlaya And how is the condition $k^2<x<(k+1)^2$ helping? Maybe your hint was also wrong, not only difficult to understand. $\endgroup$ – Did Jan 22 '17 at 9:05
  • $\begingroup$ @Did No, that's not a condition but a case analysis $\endgroup$ – kishlaya Jan 22 '17 at 9:06
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HINT

Use $\cos x - \cos y = -2 \sin(\frac {(x - y)} 2 ) \sin(\frac {(x + y)} 2 )$

Then $\lim _{x\to \infty }\sin(\frac {(\sqrt x - \sqrt {x-1})} 2 )=\lim _{x\to \infty }\sin(\frac 1 {2(\sqrt x + \sqrt {x-1})})=0$

Because $\sin(\frac {(\sqrt x + \sqrt {x-1})} 2) $ is bounded, it follows the limit is zero

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$$\cos\sqrt{x}-\cos\sqrt{x-1}=2\sin\frac{\sqrt{x}+\sqrt{x-1}}{2}\sin\frac{\sqrt{x-1}-\sqrt{x}}{2}=$$ $$=-2\sin\frac{\sqrt{x}+\sqrt{x-1}}{2}\sin\frac{1}{2(\sqrt{x-1}+\sqrt{x})}\rightarrow0$$

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The difference between $\sqrt{x}$ and $\sqrt{x-1}$ becomes arbitrary small as $x$ goes to infinity. Since $\cos$ is a uniformly continuous function, the difference between the cosine of two arbitrary close numbers is 0. So the limit is 0.

Uniform continuity is a stronger form of continuity. Having a bounded derivative is a sufficient condition for a function to be uniformly continuous.

This method requires some extra work to make a rigid proof, but it is a general method that works in many cases and it complies with intuition.

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    $\begingroup$ This answer, which provides the true insight why the limit is zero, is heavily downvoted, while at least two other answers, which rush to mostly irrelevant but explicit computations, are upvoted. Sorry but this is not something the site should be proud of. $\endgroup$ – Did Jan 21 '17 at 13:12
  • $\begingroup$ @Did I agree. People downvoted without even bother to read the answer. $\endgroup$ – egreg Jan 21 '17 at 14:47
  • $\begingroup$ @egreg Yes. Fortunately the tide of votes was somewhat reversed since then. $\endgroup$ – Did Jan 21 '17 at 15:24
  • $\begingroup$ @Did Also the content of this answer has been changed heavily $\endgroup$ – user261263 Jan 21 '17 at 15:37
  • $\begingroup$ @EugenCovaci Yeah, and it was corrected five minutes later. Do you really think the downvotes accumulated during these five minutes? First, the only downvote remaining falls outside of this time span and second, that would be an excellent news if every instant voter on the site was fully aware of the mathematical reason why the first version was lacking an argument while the second version and the subsequent ones are not. $\endgroup$ – Did Jan 22 '17 at 9:12
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$\displaystyle \lim_{x\rightarrow \infty}\bigg|\int^{\sqrt{x}}_{\sqrt{x-1}}\sin xdx\bigg| \leq \lim_{x\rightarrow \infty}\int^{\sqrt{x}}_{\sqrt{x-1}}1dx = \lim_{x\rightarrow \infty}\left(\sqrt{x}-\sqrt{x-1}\right) = \lim_{x\rightarrow \infty}\frac{1}{\left(\sqrt{x}+\sqrt{x-1}\right)}=0$

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    $\begingroup$ This proves that the $\limsup$ is $\le 0$. What about showing $\liminf \ge 0$? $\endgroup$ – Crostul Jan 21 '17 at 12:16
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    $\begingroup$ You can use absolute value around the first integral. $\endgroup$ – egreg Jan 21 '17 at 14:46
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Direct approach:

Let $f(x) := \cos\sqrt{x} - \cos\sqrt{x - 1}$.

By the mean value theorem, for two values $a, b \in \mathbb R$, there exists $t\in\mathbb [a, b]$ such that $\cos(b) - \cos(a) = \cos'(t)(b - a)$. Thus,$$ |f(x)|=|\cos\sqrt{x} - \cos\sqrt{x - 1}| \\ =|\cos'(t)(\sqrt{x} - \sqrt{x - 1})| \\ =|-\sin(t)|\cdot|\sqrt{x} - \sqrt{x - 1}|\\ \le|\sqrt{x} - \sqrt{x - 1}| = \sqrt{x} - \sqrt{x - 1} =: g(x)$$

for all $x>1$.

Assertion: $\lim_{x\rightarrow\infty} g(x) = 0$.

Proof: Let $\epsilon > 0$. Then

$$g(x) \le \epsilon\\ \Leftrightarrow\sqrt{x} \leq \epsilon + \sqrt{x - 1}\\ \Leftrightarrow x \leq \epsilon^2 + 2\sqrt{x - 1} + x -1\\ \Leftrightarrow x \geq 1+(1-\epsilon^2)^2/4 =: x_0$$

Thus, for all $x> x_0$, we have $|g(x)| = g(x) \le \epsilon$.

Together with $|f(x)|\le g(x)$, this yields $\lim_{x\rightarrow\infty} f(x) = 0$.

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  • $\begingroup$ This is the elaborated version of @Paul's answer math.stackexchange.com/a/2107266/408656 $\endgroup$ – ThomasR Jan 21 '17 at 12:49
  • $\begingroup$ I just realized that $g(x) = 1/(\sqrt{x}+\sqrt{x-1}) \rightarrow 0$, so the second part is more or less obsolete. $\endgroup$ – ThomasR Jan 21 '17 at 13:52
  • $\begingroup$ @egreg Thanks, you're right $\endgroup$ – ThomasR Jan 21 '17 at 14:49

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