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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, 2017 at 11:48
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    $\begingroup$ @kishlaya Sorry but your "hint" is not understandable. $\endgroup$
    – Did
    Jan 21, 2017 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, 2017 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, 2017 at 9:05
  • $\begingroup$ @Did No, that's not a condition but a case analysis $\endgroup$
    – kishlaya
    Jan 22, 2017 at 9:06

5 Answers 5

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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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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, 2017 at 13:12
  • $\begingroup$ @Did I agree. People downvoted without even bother to read the answer. $\endgroup$
    – egreg
    Jan 21, 2017 at 14:47
  • $\begingroup$ @egreg Yes. Fortunately the tide of votes was somewhat reversed since then. $\endgroup$
    – Did
    Jan 21, 2017 at 15:24
  • $\begingroup$ @Did Also the content of this answer has been changed heavily $\endgroup$
    – user261263
    Jan 21, 2017 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, 2017 at 9:12
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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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$\displaystyle \lim_{x\rightarrow \infty}\bigg|\int^{\sqrt{x}}_{\sqrt{x-1}}\sin xdx\bigg| \leq \lim_{x\rightarrow \infty}\int^{\sqrt{x}}_{\sqrt{x-1}}1dx = \lim_{x\rightarrow \infty}\left(\sqrt{x}-\sqrt{x-1}\right) = \lim_{x\rightarrow \infty}\frac{1}{\left(\sqrt{x}+\sqrt{x-1}\right)}=0$

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    $\begingroup$ This proves that the $\limsup$ is $\le 0$. What about showing $\liminf \ge 0$? $\endgroup$
    – Crostul
    Jan 21, 2017 at 12:16
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    $\begingroup$ You can use absolute value around the first integral. $\endgroup$
    – egreg
    Jan 21, 2017 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, 2017 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, 2017 at 13:52
  • $\begingroup$ @egreg Thanks, you're right $\endgroup$
    – ThomasR
    Jan 21, 2017 at 14:49

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