Floored Limit

Calculus Level 3

We know that by squeeze theorem , $\displaystyle \lim_{x\to0} \dfrac {\sin x}x = 1$ .

But what is is the value of $\displaystyle \lim_{x\to0} \left \lfloor \dfrac {\sin x}x \right \rfloor$ ?

0 Limit does not exist 1 -1

2 solutions

展豪張
Apr 9, 2016

$\displaystyle \lim_{x\to0}\lfloor \dfrac {\sin x}x \rfloor =\lim_{x\to0}\lfloor\dfrac{x-x^3/6+x^5/120-\cdots}x\rfloor=\lim_{x\to0}\lfloor1-x^2/6+x^4/120-\cdots\rfloor$
As it is the series expansion at $0$ , in the neighborhood of $0$ , $|\dfrac{x^4}{120}-\cdots|<|\dfrac{x^2}{6}|$ , so $1-x^2/6<1$
The answer is $0$

Would it be in any way justifiable to say that when taking the limit of this function as x approaches 0, the numerator always becomes less than the denominator after passing pi/2 causing the fraction to be 0? (Therefore the limit as well)

Andrew Tawfeek - 5 years, 2 months ago

You are absolutely correct(I think)XD

展豪張 - 5 years, 2 months ago

Edwin Gray
May 4, 2019

sin(x)/x = [x - x^3/3! + x^5/5! -......]/x. = 1 - x^2/3! +....... So in a neighborhood of 0+, sin(x)/x < 1, so the floor is 0.

To clarify: this equation only works if $|x| \leq 1$ .

Pi Han Goh - 2 years, 1 month ago

Floored Limit

2 solutions

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