Calculate the limit cos(x)/sin(x) when x approaches 0

Hi,

I need to check whether the limit of the following function exists or not
I have now proceeded as follows to look at the right-sided and left-sided limit i.e. ##\displaystyle{\lim_{x \to 0^{+}}}## and ##\displaystyle{\lim_{x \to 0^{-}}}##

To do this, I drew up a list in which I move from 1 closer and closer to 0 and for the left-hand side from -1 towards 0 and got the following:



As you can see, the function tends from the right towards ##\infty## and from the left towards ##- \infty## As the two values are therefore not equal, the function has no limit at the point ##x=0##

My question, can I prove it this way, or is there a way to prove it more precisely mathematically?

Lambda96 said:

Homework Statement: Calculate ##\displaystyle{\lim_{x \to 0}} \frac{\cos(x)}{\sin(x)}##
Relevant Equations: none

Hi,

I need to check whether the limit of the following function exists or not
View attachment 338389I have now proceeded as follows to look at the right-sided and left-sided limit i.e. ##\displaystyle{\lim_{x \to 0^{+}}}## and ##\displaystyle{\lim_{x \to 0^{-}}}##

To do this, I drew up a list in which I move from 1 closer and closer to 0 and for the left-hand side from -1 towards 0 and got the following:

View attachment 338390

As you can see, the function tends from the right towards ##\infty## and from the left towards ##- \infty## As the two values are therefore not equal, the function has no limit at the point ##x=0##

My question, can I prove it this way, or is there a way to prove it more precisely mathematically?

If you're asked to prove that the limit exists or doesn't exist, calculating a table of values doesn't do the job. What you would need to do in this case is a sort of modified ##\delta - \epsilon## argument. In this case you would need to show that for any given (large number) M, there is a (small number) ##\delta > 0## such that whenever ##|x - 0| < \delta##, where ##x \ne 0##, then ##|\frac{\cos(x)}{\sin(x)}| > M##. As a side note, ##\frac{\cos(x)}{\sin(x)}= = \cot(x)##.

Still, it seems you're working on the extended Reals, if ##\mathbb R## with the line on top *. Edit: Then the limit of infinity could exist.

*Sorry, don't know how to Latex it in.

WWGD said:

Still, it seems you're working on the extended Reals, if ##\mathbb R## with the line on top *. Then the limit exists and it's infinity.

*Sorry, don't know how to Latex it in.

##\overline{\mathbb R}##

WWGD said:

Still, it seems you're working on the extended Reals, if ##\mathbb R## with the line on top *. Then the limit exists and it's infinity.

*Sorry, don't know how to Latex it in.

That is true for the projectively extended reals. The normal extended reals have two infinities and no limit here.

Office_Shredder said:

That is true for the projectively extended reals. The normal extended reals have two infinities and no limit here.

Yes, not this expression, but either of ## \infty, -\infty ## are possible limits in
##\overline {\mathbb R}##

Thank you Mark44, WWGD, Orodruin and Office_Shredder for your help

I have now proceeded as Mark44 described:

The following applies ##|x-0|< \delta \rightarrow x < \delta## and ##\frac{\cos(x)}{\sin(x)}>M \rightarrow \cot(x)>M##

Then I calculated the following

##\cot(x) > M \qquad | \text{Form reciprocal value}##
##\frac{1}{\cot(x)} < \frac{1}{M} \qquad |\frac{1}{\cot(x)}= \tan(x)##
##\tan(x) < \frac{1}{M} \qquad | \arctan(...)##
##x < \arctan(\frac{1}{M})##

It then follows that for ##\cot(x) > M##, ##\delta## must be chosen as follows ##\delta = \arctan(\frac{1}{M})##

Lambda96 said:

Calculate ##\displaystyle{\lim_{x \to 0}} \frac{\cos(x)}{\sin(x)}##

You can use Taylor series ...

##cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-...##
##sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-...##

... and approximate with first members of two series ...
(the same as replacing functions cos(x) and sin(x) with their tangent straight lines for x=0)

##\displaystyle{\lim_{x \to 0}} \frac{\cos(x)}{\sin(x)}
=\displaystyle{\lim_{x \to 0}} \frac{1}{x}##

From the left side ##x \to -0## you can get
##\displaystyle{\lim_{x \to -0}} \frac{1}{x}
=\displaystyle{\lim_{x \to -\infty}} \frac{1}{ \frac{1}{x}}
=\displaystyle{\lim_{x \to -\infty}} x =-\infty##

Also from the right side ##x \to +0## you can get
##\displaystyle{\lim_{x \to +0}} \frac{1}{x}
=\displaystyle{\lim_{x \to +\infty}} \frac{1}{ \frac{1}{x}}
=\displaystyle{\lim_{x \to +\infty}} x =+\infty##

Cotangent function is ##ctg(x)=cot(x)=\frac{\cos(x)}{\sin(x)}##
Wolfram alpha -> https://www.wolframalpha.com/input?i=cos(x)/sin(x)

You can see that [itex]\cos x/\sin x[/itex] is odd, so either the limit is zero or the limit does not exist. The limit is not zero: since [itex]\tan x \to 0[/itex], for each [itex]R > 0[/itex] there exists [itex]\delta > 0[/itex] such that if [itex]0 < |x| < \delta[/itex] then [itex]|\tan x| < R^{-1}[/itex] and [itex]|\cot x| > R[/itex].

I'm not convinced you need an epsilon-delta proof, as long as you can use basic properties of sine and cosine.
$$\lim_{x \to 0} \cos x =1$$$$\lim_{x \to 0}\sin x = 0$$$$0 < x < \pi \ \Rightarrow \ \sin x > 0$$$$-\pi <x<0 \ \Rightarrow \ \sin x < 0$$

Lambda96 said:

As you can see, the function tends from the right towards ##\infty## and from the left towards ##- \infty## As the two values are therefore not equal, the function has no limit at the point ##x=0##

My question, can I prove it this way, or is there a way to prove it more precisely mathematically?

This is correct. You have shown the limit does not exist.

In the topology of the Extended Reals, the limit as ##\infty## may exist if your expression grows without bound in the positive direction, i.e., it would be in the neighborhood ##(a, \infty]; a>0##. Similar for ##-\infty ## as a limit. That's not the case here, as cotan alternates signs.
https://en.m.wikipedia.org/wiki/Extended_real_number_line

Thank you Bosko, pasmith, PeroK, nuuskur and WWGD for your help

Thanks PeroK with the tip to argue that the sine around ##x=0## is asymmetric and therefore the limit ##\displaystyle{\lim_{x \to 0^-}} \frac{\cos(x)}{\sin(x)}=- \infty## and ##\displaystyle{\lim_{x \to 0^+}} \frac{\cos(x)}{\sin(x)}= \infty## and therefore the limit ##\displaystyle{\lim_{x \to 0}} \frac{\cos(x)}{\sin(x)}## does not exist