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

[Lambda96]

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

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?

 

[Mark44]

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, $\frac{\cos(x)}{\sin(x)}= = \cot(x)$.

 

[WWGD]

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.

 

[Orodruin]

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.

$\overline{\mathbb R}$

 

[Office_Shredder]

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.

 

[WWGD]

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
$\overline {\mathbb R}$

 

[Lambda96]

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 $\delta = \arctan(\frac{1}{M})$

 

[Bosko]

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)

 

[pasmith]

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

 

[PeroK]

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

 

[nuuskur]

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.

 

[WWGD]

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

 

[Lambda96]

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