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I was doing this question, i think the answer should be $0$ ,

My argument is as follows suppose $x \rightarrow 0$ from right side, that is $x $ approaches $0$ from positive side meaning $x$ is decreasing and therefore $\cos(x) $ should increase as $\cos(x)$ is a decreasing function , but $\cos(x) < 1$ hence their integral part will be $0$ , similarly for the negative case the $\cos(x)$ plays the same role as $\cos(x)$ is an even function.

but at the same time i doubt what happens when $x = 0$,there $\cos(x) = 1$ and hence integer part will be $1$.

Please correct me if i am wrong.

  • 1
    I guess 0 is the answer because limit is in the neighbourhood of the point not at the point. – 2017-02-18
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    You have that $[\cos x]=0$ for all $x \ne 0$ in an open interval about $0,$ so in effect you are evaluating the limit $\lim_{x \to 0} 0.$ – 2017-02-18

2 Answers 2

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You're right.

The function $f(x)=\lfloor \cos x\rfloor$ is constant in a punctured neighborhood of $0$: for $x\in(-\pi/2,\pi/2)$, $x\ne0$, $$ \lfloor \cos x\rfloor=0 $$ because $0<\cos x<1$. Therefore $$ \lim_{x\to0}\lfloor \cos x\rfloor=0 $$

What the value of $\lfloor \cos x\rfloor$ is for $x=0$ is completely irrelevant as far as the limit is concerned, except for the fact you can conclude that $f$ is not continuous at $0$.

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You're basically right. The actual value of $\lfloor \cos x \rfloor$ at $x=0$ is irrelevant: limits talk about values in a neighbourhood and not at a point.

$\lfloor \cos x \rfloor = 0$ for $0 < x < \frac{1}{2}$ (with room to spare!), so the value of $\lfloor \cos \frac{1}{n} \rfloor$ is equal to $0$ for all $n \geq 2$. Therefore, if the limit as $x \to 0$ exists, it is equal to $0$ (because the limit is $0$ if we approach it in this special way, along the sequence $\frac{1}{i}$).