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How are the two expressions different?

$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}}{x}\bigg\rfloor$$

and $$\bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor$$

If limit is inside the floor function, Do I have to apply the limits first?

If this is the case, then, $$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}}{x}\bigg\rfloor=0$$

$$\bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor=1$$

Am I solving this right?

Also how can I calculate,

$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor$$

Thank you.

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  • $\begingroup$ You're correct; Note that $0\le \cos(x)\le \frac{\sin(x)}{x}\le 1$ for $0<|x|< \pi/2$. $\endgroup$ – Mark Viola Aug 29 '18 at 15:22
  • $\begingroup$ @postmortes ${\sin x\over x}$ is never negative near $0$ is it? How could the floor be $-1?$ $\endgroup$ – saulspatz Aug 29 '18 at 15:23
  • $\begingroup$ Just a quick comment: If $\lfloor x\rfloor$ was a continuous function, then you would be able to swap it with $\lim$. $\endgroup$ – Jason DeVito Aug 29 '18 at 17:16
  • $\begingroup$ Partly similar: math.stackexchange.com/questions/2491184/… $\endgroup$ – Hans Lundmark Aug 29 '18 at 19:11
  • $\begingroup$ @prog_SAHIL Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – user Sep 17 '18 at 20:24
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$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}}{x}\bigg\rfloor$$ In this first, you have to take the floor of the function then apply the limit on its floor. $$\frac{\sin{x}}{x}< 1$$ when $x \to 0$ $$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}}{x}\bigg\rfloor=0$$

$$\bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor$$

Here you have to first calculate the limit then take the floor of it. You know that this limit is 1 thus floor of 1 is 1. $$\bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor=1$$

$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor$$

Series expansion at $x=0$ gives $$1+\frac{x^2}{6}+\frac{31x^4}{360}+O(x^6)$$

$$\frac{\sin{x}\cdot \tan{x}}{x^2}\ge 1$$

$$\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor=1$$ thus $$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor=1$$

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The first expression says:

  • Take the function $f(x) = \lfloor\frac{\sin(x)}{x}\rfloor$.
  • Take the limit $$\lim_{x\to 0} f(x)$$

The second expression says:

  • Take the function $f(x) = \frac{\sin(x)}{x}$.
  • Take the limit $$L=\lim_{x\to 0} f(x)$$.
  • Calculate $\lfloor L\rfloor$

The two expressions do not need to be the same.

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Yes you are right indeed since

$$\frac{\sin x}{x}<1 \implies \bigg \lfloor\frac{\sin{x}}{x}\bigg\rfloor=0 \implies \lim_{x\to0} \bigg \lfloor\frac{\sin{x}}{x}\bigg\rfloor=0$$

and since

$$\lim_{x\to0} \frac{\sin{x}}{x}=1 \implies \bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor=1 $$

For the latter for $x$ sufficiently small we have

$$1<\frac{\sin{x}\cdot \tan{x}}{x^2}<2$$

indeed $\sin x>x-\frac{x^3} 6$ and $\tan x>x+\frac{x^3} 3$ and

$$\frac{\sin{x}\cdot \tan{x}}{x^2}>1+\frac{x^2}6-\frac{x^6}{18}$$

therefore

$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor=1$$

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Here's fact 4

If $f(x)$ is continuous at $a$ and $\mathop {\lim }\limits_{x \to a} g\left( x \right) = b$ then:

$\mathop {\lim }\limits_{x \to a} f\left( {g\left( x \right)} \right) = f\left( {\mathop {\lim }\limits_{x \to a} g\left( x \right)} \right)$

Floor function isn't continuous, so you should apply floor function first, before calculating limit.

Let's calculate

$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor$$

Okay, the first thing I want to do it's to solve: $$\lim_{x\to0}\frac{\sin{x}\cdot \tan{x}}{x^2} = \lim_{x\to0} \frac{\sin{x}}{x}{\frac{\sin{x}}{x}}{\frac{1}{\cos{x}}} = [\lim_{x\to0}\frac{\sin{x}}{x} = 1] = \lim_{x\to0}\frac{1}{\cos{x}} = 1$$

So, $$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor = \lim_{x\to0}\bigg\lfloor\frac{1}{\cos{x}}\bigg\rfloor $$ $$\cos{x} \le 1 $$$$\frac{1}{\cos{x}} \ge 1$$

$$x < \arccos(2) => \cos{x} < \frac{1}{2} => \frac{1}{\cos{x}}< 2$$

It means that in the neighborhood of 0 with radius $\delta < arccos(2)$ function $\bigg\lfloor\frac{1}{\cos{x}}\bigg\rfloor \equiv 1$

And we can conclude that

$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor = \lim_{x\to0}\bigg\lfloor\frac{1}{\cos{x}}\bigg\rfloor = 1 $$

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