# Trig limit 1

#### Petrus

##### Well-known member
Hello,
I got problem solving $$\lim_{x \to 0}\frac{\cos(x)-1}{\sin(x)}$$
I do not really get any progress, I would be glad if someone could give me tips!

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Trig limit

There are many ways to solve it :

you may use : $$\displaystyle \cos(2x)= 1-2\sin^2(x)$$

you may also differentiate both numerator and denominator (LH rule)

series expansion ...

#### Bacterius

##### Well-known member
MHB Math Helper
Re: Trig limit

Are you familiar with l'Hôpital's rule, or are you meant to find the limit a particular way?

#### Petrus

##### Well-known member
Re: Trig limit

No I do not wanna use l hopitals rule. I could not understand that tips u gave me with cos(2x).
basicly I think ima divide and multiplying. Thats what My book have done on example and i read on this chapter

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Trig limit

Since we have the double angle identity :

$$\displaystyle \cos(2x)= 1-2\sin^2(x)$$

so $$\displaystyle \cos(2x)-1= -2\sin^2(x)$$

But there we have cos(x) so Let us generalize

$$\displaystyle \cos(angle) -1 = -2\sin^2(angle/2)$$

so we have $$\displaystyle \cos(x)-1= -2\sin^2\left(\frac{x}{2}\right)$$

#### chisigma

##### Well-known member
Re: Trig limit

Hello,
I got problem solving $$\lim_{x \to 0}\frac{\cos(x)-1}{\sin(x)}$$
I do not really get any progress, I would be glad if someone could give me tips!
Setting $\displaystyle x= 2\ \theta$ and considering that is...

$\sin 2\ \theta = 2\ \sin \theta\ \cos \theta$

$\cos 2\ \theta = 1 - 2\ \sin^{2} \theta$ (1)

... You arrive to write...

$\displaystyle \lim_{x \to 0}\frac{\cos(x)-1}{\sin(x)}= \lim_{\theta \rightarrow 0} - \frac{\sin \theta}{\cos \theta} =0$ (2)

Kind regards

$\chi$ $\sigma$

#### Petrus

##### Well-known member
Re: Trig limit

Hello,
Does it work if I use $\cos^2(x)+\sin^2(x)=1$ and rewrite as $\cos^2(x)-1=-\sin^2(x)$
and at last step I get
$$\lim_{x \to 0} \frac{-\sin(x)}{\cos(x)+1}=-0/2=0$$.

Note that i simplifie before last step

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#### Petrus

##### Well-known member
Re: Trig limit

Since we have the double angle identity :

$$\displaystyle \cos(2x)= 1-2\sin^2(x)$$

so $$\displaystyle \cos(2x)-1= -2\sin^2(x)$$

But there we have cos(x) so Let us generalize

$$\displaystyle \cos(angle) -1 = -2\sin^2(angle/2)$$

so we have $$\displaystyle \cos(x)-1= -2\sin^2\left(\frac{x}{2}\right)$$
Hello ZaidAlyafey,
If you possible got time I would be glad if you could show me step by step on your way. I would really be glad if I could solve this problem on diffrent way (I could unfortenetly not solve with your way)

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Trig limit

Ok , I will use the facts :

$$\displaystyle \cos(x)-1=-2\sin^2\left(\frac{x}{2}\right)$$

$$\displaystyle \sin(x) = 2\sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)$$

So substituting this in our limit we get :

$$\displaystyle \lim_{x\to 0}\frac{-2\sin^2\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)}=\lim_{x \to 0}\frac{-\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)}=\frac{0}{1}=0$$

#### Petrus

##### Well-known member
Re: Trig limit

Ok , I will use the facts :

$$\displaystyle \cos(x)-1=-2\sin^2\left(\frac{x}{2}\right)$$

$$\displaystyle \sin(x) = 2\sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)$$

So substituting this in our limit we get :

$$\displaystyle \lim_{x\to 0}\frac{-2\sin^2\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)}=\lim_{x \to 0}\frac{-\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)}=\frac{0}{1}=0$$
Thanks!
(This was My bad... I was not thinking correct.. I was just rewriting cos and not sin.... Darn next time I have to think about this...)

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#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Trig limit

Hello,
Does it work if I use $\cos^2(x)+\sin^2(x)=1$ and rewrite as $\cos^2(x)-1=-\sin(x)$
and at last step I get
$$\lim_{x \to 0} \frac{-\sin(x)}{\cos(x)+1}=-0/2=0$$.

Note that i simplifie before last step
No , you are missing a square , $\cos^2(x)-1=-\sin^2(x)$

#### Petrus

##### Well-known member
Re: Trig limit

No , you are missing a square , $\cos^2(x)-1=-\sin^2(x)$
Thanks! Forgot when i wrote it on pc!

#### MarkFL

Staff member
Re: Trig limit

Hello,
I got problem solving $$\lim_{x \to 0}\frac{\cos(x)-1}{\sin(x)}$$
I do not really get any progress, I would be glad if someone could give me tips!
I would rewrite the expression as follows:

$$\displaystyle \frac{\cos(x)-1}{\sin(x)}\cdot\frac{\cos(x)+1}{\cos(x)+1}=\frac{\cos^2(x)-1}{\sin(x)(\cos(x)+1)}=\frac{-\sin^2(x)}{\sin(x)(\cos(x)+1)}=-\frac{\sin(x)}{\cos(x)+1}$$

and so:

$$\displaystyle \lim_{x \to 0}\frac{\cos(x)-1}{\sin(x)}=-\lim_{x \to 0}\frac{\sin(x)}{\cos(x)+1}=-\frac{0}{1+1}=0$$

This is essentially what you did when you tried using a Pythagorean identity, but I wanted to clearly lay out all the steps.

#### Chris L T521

##### Well-known member
Staff member
Re: Trig limit

Hello,
I got problem solving $$\lim_{x \to 0}\frac{\cos(x)-1}{\sin(x)}$$
I do not really get any progress, I would be glad if someone could give me tips!
There is yet another way to do this one if you're allowed to use the facts that $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1$ and $\displaystyle\lim_{x\to 0}\frac{1-\cos x}{x}=0$.

Then it follows that
$\lim_{x\to 0}\frac{\cos x -1 }{\sin x}=\lim_{x\to 0} \frac{-\dfrac{1-\cos x}{x}}{ \dfrac{\sin x}{x}}= -\frac{\displaystyle\lim_{x\to 0} \frac{1-\cos x}{x}}{ \displaystyle \lim_{x\to 0}\frac{\sin x}{x}} =-\frac{0}{1}=0$

I hope this makes sense!

#### Siron

##### Active member
Re: Trig limit

You can also work with taylorseries.
We have
$\sin(x) = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots$
$\cos(x) = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots$

We obtain
$\displaystyle \lim_{x \to 0} \frac{\cos(x)-1}{\sin(x)} = \displaystyle \lim_{x \to 0} \frac{\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots\right)-1}{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots} = \lim_{x \to 0} \frac{-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots}{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots}$
$=\displaystyle \lim_{x \to 0} \frac{x \left(-\frac{x}{2!}+\frac{x^3}{4!}-\frac{x^5}{6!}+\ldots\right)}{x \left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\ldots\right)} = \frac{0}{1}= 0$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Trig limit

Are we running out of methods ...

#### MarkFL

Staff member
Re: Trig limit

Are we running out of methods ...
Well, there is still L'Hôpital's rule, but Petrus has stated he wished not to use this method. #### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: Trig limit

Are we running out of methods ...
We could still try the squeeze rule...
Also pretty standard for this type of problem.

#### Petrus

##### Well-known member
Re: Trig limit

You can also work with taylorseries.
We have
$\sin(x) = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots$
$\cos(x) = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots$

We obtain
$\displaystyle \lim_{x \to 0} \frac{\cos(x)-1}{\sin(x)} = \displaystyle \lim_{x \to 0} \frac{\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots\right)-1}{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots} = \lim_{x \to 0} \frac{-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots}{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots}$
$=\displaystyle \lim_{x \to 0} \frac{x \left(-\frac{x}{2!}+\frac{x^3}{4!}-\frac{x^5}{6!}+\ldots\right)}{x \left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\ldots\right)} = \frac{0}{1}= 0$
This method have to wait Havent read about Taylor yet #### Petrus

##### Well-known member
Re: Trig limit

Hello,
In exam would you guys prefer lhopitals rule? Well for me it seems like A easy trick. I try always solve with lhopitals rule and without, dont wanna get stuck with just going for lhopitals.

(Edit: I wanna Clear that i am not saying that derivate is simple but most of limits of My book dont have really have difficoult to derivate. I wanna also say yes i know to be able to use lhopitals i have to check if original function is $\frac{0}{0}$ $\frac{\infty}{\infty}$ $\frac{-\infty}{-\infty}$ and yes i know My infinity latex code does not work and i do not know how to get it correct)

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: Trig limit

If you're allowed to use l'Hopital in an exam why not use it?
To get infinity in latex use \infty ($\infty$).

As for yet another way to get the limit with the squeeze theorem...

If $x > 0$ we have:
$$- \frac {x^2} 2 \le \cos x - 1 \le 0$$
$$0 \le \sin x \le x$$

It follows that:
$$\frac{- \frac {x^2} 2}{x} \le \frac{\cos x - 1}{\sin x} \le \frac 0 {\sin x}$$
$$- \frac {x} 2 \le \frac{\cos x - 1}{\sin x} \le 0$$

Since the left and right hand side both approach zero if x approaches zero, it follows by the squeeze theorem that:
$$\lim_{x \downarrow 0} \frac{\cos x - 1}{\sin x} = 0$$

Similarly we can find that:
$$\lim_{x \uparrow 0} \frac{\cos x - 1}{\sin x} = 0$$

So it follows that:
$$\lim_{x \to 0} \frac{\cos x - 1}{\sin x} = 0 \qquad \blacksquare$$

#### MarkFL

Staff member
Re: Trig limit

It has been my experience that in the vast majority of cases, L'Hôpital's rule is the most straightforward way to handle indeterminate forms. Even if you are instructed to use another method, it is a good tool to use to check your results.

#### chisigma

##### Well-known member
Re: Trig limit

It has been my experience that in the vast majority of cases, L'Hôpital's rule is the most straightforward way to handle indeterminate forms. Even if you are instructed to use another method, it is a good tool to use to check your results.
Unfortunately there are unfrequent cases in which l'Hopital rule is a wrong way... for example let's suppose to have the following limit...

$\displaystyle \lim_{x \rightarrow 0} \frac{x^{2}\ \sin \frac{1}{x}}{\sin x}$ (1)

Applying l'Hopital rule You arrive to the expression...

$\displaystyle \frac {2\ x\ \sin \frac{1}{x} - \cos \frac{1}{x}}{\cos x}$ (2)

... which has no limit if x tends to 0. If You proceed 'directly' however You obtain...

$\displaystyle \lim_{x \rightarrow 0}\ \frac{x^{2}\ \sin \frac{1}{x}}{\sin x} = \displaystyle \lim_{x \rightarrow 0}x\ \frac{x}{\sin x}\ \sin \frac{1}{x} = 0$ (3)

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
Re: Trig limit

Unfortunately there are unfrequent cases in which l'Hopital rule is a wrong way... for example let's suppose to have the following limit...

$\displaystyle \lim_{x \rightarrow 0} \frac{x^{2}\ \sin \frac{1}{x}}{\sin x}$ (1)

Applying l'Hopital rule You arrive to the expression...

$\displaystyle \frac {2\ x\ \sin \frac{1}{x} - \cos \frac{1}{x}}{\cos x}$ (2)

... which has no limit if x tends to 0. If You proceed 'directly' however You obtain...

$\displaystyle \lim_{x \rightarrow 0}\ \frac{x^{2}\ \sin \frac{1}{x}}{\sin x} = \displaystyle \lim_{x \rightarrow 0}x\ \frac{x}{\sin x}\ \sin \frac{1}{x} = 0$ (3)
Of course the example I reported could be given to a student in an exam test and in that case the reccommendation don't to use l'Hopital test would be essential. But does that type of 'decency' ever used by teachers?...

Kind regards

$\chi$ $\sigma$

#### Petrus

##### Well-known member
Re: Trig limit

Unfortunately there are unfrequent cases in which l'Hopital rule is a wrong way... for example let's suppose to have the following limit...

$\displaystyle \lim_{x \rightarrow 0} \frac{x^{2}\ \sin \frac{1}{x}}{\sin x}$ (1)

Applying l'Hopital rule You arrive to the expression...

$\displaystyle \frac {2\ x\ \sin \frac{1}{x} - \cos \frac{1}{x}}{\cos x}$ (2)

... which has no limit if x tends to 0. If You proceed 'directly' however You obtain...

$\displaystyle \lim_{x \rightarrow 0}\ \frac{x^{2}\ \sin \frac{1}{x}}{\sin x} = \displaystyle \lim_{x \rightarrow 0}x\ \frac{x}{\sin x}\ \sin \frac{1}{x} = 0$ (3)

Kind regards

$\chi$ $\sigma$
Hello,
Sorry making this many page but i never knew about this. Why does this happend? I know for example equation you can get fake root but what is the reason here?