@robjohn but above is the fractional of x as well

$\lim\limits_{{x}\to0}\frac{x}{\tan(x)}$

so sorry sir for replying late actually I am in the rural area so network connectivity is not good here

I am not sure what you mean. $\frac{x}{\tan(x)}=\frac{\cos(x)}{\frac{\sin(x)}x}$

$\lim\limits_{{x}\to0}\frac{{x}}{\tan(x)}$

sir x in numerator and denominator is the fractional part of x replace x by fractional part of x sir

Well, then write that. Do you mean $\{x\}=x-\lfloor x\rfloor$, or the odd function $\mathrm{sign}(x)\{|x|\}$?

the first one

$\{x\}=x-\lfloor x\rfloor$

then the limit is not $1$

why sir?

from the right it is $1$. From the left it is $\cot(1)$

$\lim\limits_{x\to0}\frac{\{x\}}{\tan(\{x\})}$ does not exist.

can we prove it anyaltically

from the left $\lim\limits_{x\to0^-}\{x\}=-1$

how?

That was the definition of $\{x\}=x-\lfloor x\rfloor$ you chose.

how can this be possible where the exact value of the function can't be 1 or -1 how can it's limit equal to that

$\lfloor-0.0001\rfloor=-1$ so $\{-0.0001\}=-0.0001-\lfloor-0.0001\rfloor=0.9999$

yes

but that is an exact value somewhere on the line?

what do you mean? Those are exact values.

sorry sir nothing I got it one more similar question sir if x tends to infinity then?h

$\{-0.0000001\}=0.9999999$

$\lim\limits_{x\to\infty}\{x\}$ does not exist.

by the same logic?

as $x\to\infty$, $\{x\}$ fluctuates between $0$ and $1$

yes

like a sawtooth wave

/|/|/|

yes sir

and tanx will also be near zero

@robjohn i was solving one of the questions where he says $\lim\limits_{x\to\infnty}\lfloor x\rfloor$

same limit x tends to infinity

is it correct ot say this?

what does he say about $\lim\limits_{x\to\infty}\lfloor x\rfloor$?

yes

???

$\lim\limits_{x\to\infty}\lfloor x\rfloor=\infty$

what does he say??

he says it limits be same as of the $\lim\limits_{x\to\infty}\lfloor x\rfloor=\infty$

i mean$\lim\limits_{x\to\infty} x=\infty$

$\lim\limits_{x\to\infty}\{x\}$ does not exist.

it gif

are you not seeing the MathJax rendered still?

i can see

Okay, it seemed as if you were trying to write $\{x\}$ and were writing ${x}$ instead

$\lim\limits_{x\to\infty}x=\lim\limits_{x\to\infty}\lfloor x\rfloor=\infty$

i am lost i feel

i mean this only

$\lim\limits_{x\to\infty}\{x\}$ does not exist

@Yuvraj okay

it is a new question

Do you not understand that which you have quoted twice?

not fully

they both grow unboundedly as $x$ does.

$\lim\limits_{x\to\infty}x=\lim\limits_{x\to\infty}\lfloor x\rfloor=\infty$ but $\lim\limits_{x\to\infty}\left(x-\lfloor x\rfloor\right)$ does not exist

yes sir so it is correct?

if $x\ge100000$, then $\lfloor x\rfloor\ge100000$, etc. so if you mean $\lim\limits_{x\to\infty}\lfloor x\rfloor=\infty$, then, yes, that is correct.

ok, sir, I have one more fundamental question if I write $\lfloor {sinx+cosx}\rfloor$

how would i draw

it sir always struggle whenever I try to fraw gif graph

@robjohn

It might be more explanatory to see this...

yes

thanks for your time sir

Here is a labeled version

@robjohn sir if you are free please kindly look at this figure I need to find the maximum and minimum area of quadrilateral

Since the points on the sides of the square

maximum is $1$, minimum is $0$.

the area is $\frac12(1+d-b+ab+bc-cd-ad)$

$a=b=c=0$ and $d=1$ gives $1$

$a=c=d=1$ and $b=0$ gives $0$