Question

lim x-0 {x}/tan{x} , where {} , denotes the fractional part

Answer 1

use tanx=sinx/cosx

Answer 2

and use the fact that x/sinx goes to 1 as x goes to 0

Answer 3

Method 1: \[\Large \begin{align}\lim_{x\rightarrow0}\frac{x}{\tan x}&=\lim_{x\rightarrow0} \frac{x}{\frac{\sin x}{\cos x}} \\ ~ \\&=\lim_{x\rightarrow0} \frac{x }{\sin x}\cos x \\ ~ \\&=\lim_{x\rightarrow0} \frac{1}{\frac{\sin x}{x}}\cos x \\ ~ \\ &=\frac{1}{\lim_{x\rightarrow0} \frac{\sin x}{x}} \lim_{x\rightarrow0}\cos x\end{align} \]

Answer 4

Another method would be to use l'Hôpital's rule but not sure if you would have seen that already

Answer 5

Or use the graph ^_^

Answer 6

You can write the fractional part of \(x\) as a piecewise function near zero:
\[\{x\}=\begin{cases}
x&\text{for }x\ge0\\
x+1&\text{for }x<0\end{cases}\]
which means
\[\frac{\{x\}}{\tan x}=\begin{cases}
\dfrac{x}{\tan x}&\text{for }x\ge0\\\\
\dfrac{x+1}{\tan x}&\text{for }x<0\end{cases}\]
Then determine the one-sided limits.

Answer 7

As the others mentioned, you can rewrite as
\[\frac{\{x\}}{\tan x}=\frac{\{x\}}{\sin x}\cos x\]
For the right-sided limit, you'll get
\[\large\lim_{x\to0^+}\frac{x}{\sin x}\cos x=\left(\lim_{x\to0^+}\frac{x}{\sin x}\right)\left(\lim_{x\to0^+}\cos x\right)=1\cdot1=1\]
The left-sided limit is
\[\large\lim_{x\to0^-}\frac{x+1}{\sin x}\cos x=\frac{1}{0}=DNE\]
The limit doesn't exist.

Answer 8

Its tan{x} , not just tan x !
I think you guys missed that ..

Answer 9

Oh I must have read them as parentheses...

Anyway, you can define it as a similar piecewise function:
\[\frac{\{x\}}{\tan \{x\}}=\begin{cases}
\dfrac{x}{\tan x}&\text{for }x\ge0\\\\
\dfrac{x+1}{\tan (x+1)}&\text{for }x<0\end{cases}\]

Answer 10

So , it will open as :
RHL- lim x->0 (x/ tanx)...which is 1
LHL- lim x->0 (x+1/tan(x+1)) which is 1 / tan 1

Answer 11

The problem is answer is given 1 , but it should be limit does not exist right ?

Answer 12

No it is 1

Answer 13

\(\Large \rm\color{midnightblue}{\lim_{x\to0}\frac{x}{tan(x)}}\)
doing what kirby and siths said you get limit x/sinx times limit cosx
so 1.1=1

Answer 14

it is not just x , but fractional part of x , i agree with you for RHL , but in LHL , due to change in definition , answer changes..

Answer 15

I agree that the limit doesn't exist... Left-handed is definitely 1. The right-handed one seems to be \(\cot1\).

Answer 16

*** RIGHT-handed is definitely 1, LEFT-handed is \(\cot1\). ***

Answer 17

Plot: http://www.wolframalpha.com/input/?i=Sawtooth%5Bx%5D%2FTan%5BSawtooth%5Bx%5D%5D

Answer 18

what is saw tooth wave ??

Answer 19

I disagree hehe
i think we don't need to do left hand side and right hand side limits
as we proved that limit exist in for both left and right hand side
when we did limit x===>0 we found that it is 1 not further steps needed i guess?

Answer 20

oh i guess i miss understood what that function is?

Answer 21

yeah , ....the function is fractional part of x , it definition changes after an interval of 1 unit :)

Answer 22

the wesite shows {x}/ tanx not {x}/tan{x} :(

Answer 23

@Dexter810 hold on... I have the wrong function in mind. Apparently, it should be (for \(x\) near 0)
\[\{x\}=x~~\text{ for }-1\le x\le1\]
not \(0\le x\le1\). Sorry for the confusion. I took this class on signal processing and I remember the sawtooth wave being defined in terms of the fractional part function, and that was on my mind at first.

http://www.wolframalpha.com/input/?i=fractional+part+of+x

Forget about the sawtooth wave. Refer to @kirbykirby's post. All the math is right.

Here's a plot of the given function: http://www.wolframalpha.com/input/?i=frac%28x%29%2FTan%5Bfrac%5Bx%5D%5D

Answer 24

Oh . No problem .
Umm , I think
http://www.wolframalpha.com/input/?i=fractional+part+of+x
is wrongly ploted .
Here is the correct one ( I think ) :
http://hardycalculus.com/calcindex/IE_fractionalpart.htm

Answer 25

fractional part of x can never be negative , its defined like say ,x= 5.45 , then fractional part of x =0.45 , say for x=-5.45 , fractional part of x = 1-0.45 =0.55 ( which comes when eq is y=x+6 , hence i think the link which i sent is the correct graph .

Answer 26

Okay, so according to your link, the fractional part is exactly the same as the Sawtooth function ... >.<

According to WolframMathWorld, the fractional part function doesn't have a generally agreed upon definition. Compare the plots at the top of the page in this link: http://mathworld.wolfram.com/FractionalPart.html

You're using the Graham definition, which is says \(\{x\}=x-\lfloor x\rfloor\), which is given by \(\text{Sawtooth}(x)\) in Mathematica.

Answer 27

ha ha , i guess , it is same as sawtooth :D !
yep , [x] + {x}=x , thats the definition i have been taught , and the author of the book , assumes us to use this definition !

Answer 28

Anyways , Thanks a lot all of you guys :D

Answer 29

If that's the case, then the limit certainly doesn't exist, like I said earlier.

You're welcome!

Answer 30

Oh I definitely read that wrong. I thought you were trying to write like \frac{}{} in LaTeX. My bad :o

Answer 31

No problem :)