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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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    Hint: Consider all possible ways in which $x$ approaches $\infty$ – 2017-01-21
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    @kishlaya Sorry but your "hint" is not understandable. – 2017-01-21
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    @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. – 2017-01-22
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    @kishlaya And how is the condition $k^2 – 2017-01-22
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    @Did No, that's not a condition but a case analysis – 2017-01-22
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    @kishlaya Then, as I said, it is **wrong**. But if you want to explain what you mean by expanding this "hint" into a full answer, please do that, and if indeed your proof works, I will be delighted. – 2017-01-22

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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    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. – 2017-01-21
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    @Did I agree. People downvoted without even bother to read the answer. – 2017-01-21
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    @egreg Yes. Fortunately the tide of votes was somewhat reversed since then. – 2017-01-21
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    @Did Also the content of this answer has been changed heavily – 2017-01-21
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    @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. – 2017-01-22
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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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    This proves that the $\limsup$ is $\le 0$. What about showing $\liminf \ge 0$? – 2017-01-21
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    You can use absolute value around the first integral. – 2017-01-21
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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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    This is the elaborated version of @Paul's answer http://math.stackexchange.com/a/2107266/408656 – 2017-01-21
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    I just realized that $g(x) = 1/(\sqrt{x}+\sqrt{x-1}) \rightarrow 0$, so the second part is more or less obsolete. – 2017-01-21
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    @egreg Thanks, you're right – 2017-01-21