why is the lim_x->0 |sin(x)|/x doesn't exist

[zenox]

I have recently watched a video where someone concluded that \lim\limits_{x->0}\frac{|sin(x)|}{x} doesn't exist because

\lim\limits_{x->0^-}\frac{|sin(x)|}{x} = -1 and
\lim\limits_{x->0^+}\frac{|sin(x)|}{x} = 1.
I am aware that \lim\limits_{x->0}\frac{sin(x)}{x} but I am not sure why somehow |sin(x)| here when \lim\limits_{x->0} is assumed to be -cos(x) (if we apply l'hopital's rule) because the |sin(x)| should always be positive, it's the same as taking the values of x from the first and second quadrant.
I am sure I'm missing something here
can anyone help me with this?
Thanks

 

[Deleted member 4993]

I have recently watched a video where someone concluded that \lim\limits_{x->0}\frac{|sin(x)|}{x} doesn't exist because

\lim\limits_{x->0^-}\frac{|sin(x)|}{x} = -1 and
\lim\limits_{x->0^+}\frac{|sin(x)|}{x} = 1.
I am aware that \lim\limits_{x->0}\frac{sin(x)}{x} but I am not sure why somehow |sin(x)| here when \lim\limits_{x->0} is assumed to be -cos(x) (if we apply l'hopital's rule) because the |sin(x)| should always be positive, it's the same as taking the values of x from the first and second quadrant.
I am sure I'm missing something here
can anyone help me with this?
Thanks

Plot sin(x)/x for x = -2 to 2 and observe......

Then plot |sin(x)|/x and observe further......

 

[Steven G]

\(\displaystyle \dfrac{sin(-x)}{-x} = \dfrac{-sin(x)}{-x} = \dfrac{sin(x)}{x}\)
The above is the key.

 

[zenox]

Sorry for posting such a lame question I somehow disregarded the x in the fraction lmao.
Thank you all for your help

\(\displaystyle \dfrac{sin(-x)}{-x} = \dfrac{-sin(x)}{-x} = \dfrac{sin(x)}{x}\)
The above is the key.

Sorry for such a lame question and thank you for taking the time to response

 

[Steven G]

So does the limit exist or not, ie is the video correct?

 

[Steven G]

I just realized that you are computing the limit of |sinx|/x not sinx/x

 

[zenox]

So does the limit exist or not, ie is the video correct?

As I understand it now, it doesn't, and the video is correct.
|sinx|/x will have two different values, it's true that |sinx| will be always positive but as you pointed out the denominator is either positive or negative depending on approaching 0 either from the right or the left.
However, sinx/x will be 1, as negative values will cancel out.
Please correct me if I am wrong.

 

[Steven G]

As I understand it now, it doesn't, and the video is correct.
|sinx|/x will have two different values, it's true that |sinx| will be always positive but as you pointed out the denominator is either positive or negative depending on approaching 0 either from the right or the left.
However, sinx/x will be 1, as negative values will cancel out.
Please correct me if I am wrong.

You are correct. very good.