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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

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!

- Thread starter Petrus
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- Thread starter
- #1

- Feb 21, 2013

- 739

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!

- Jan 17, 2013

- 1,667

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 ...

- Jan 26, 2012

- 644

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

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- #4

- Feb 21, 2013

- 739

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

- Jan 17, 2013

- 1,667

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

- Feb 13, 2012

- 1,704

Setting $\displaystyle x= 2\ \theta$ and considering that is...

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!

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

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

$\chi$ $\sigma$

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- #7

- Feb 21, 2013

- 739

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

Last edited:

- Thread starter
- #8

- Feb 21, 2013

- 739

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

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)

- Jan 17, 2013

- 1,667

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 :

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

- Thread starter
- #10

- Feb 21, 2013

- 739

Thanks!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 :

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

(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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- Jan 17, 2013

- 1,667

No , you are missing a square , $\cos^2(x)-1=-\sin^2(x)$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

- Thread starter
- #12

- Feb 21, 2013

- 739

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

- Admin
- #13

I would rewrite the expression as follows:

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!

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

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.

- Moderator
- #14

- Jan 26, 2012

- 995

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 $\displaystyle\lim_{x\to 0}\frac{1-\cos x}{x}=0$.

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!

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!

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

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

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

- Jan 17, 2013

- 1,667

Are we running out of methods ...

- Admin
- #17

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

- Admin
- #18

- Mar 5, 2012

- 8,899

We could still try the squeeze rule...Are we running out of methods ...

Also pretty standard for this type of problem.

- Thread starter
- #19

- Feb 21, 2013

- 739

This method have to wait Havent read about Taylor yetYou 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

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

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

- Thread starter
- #20

- Feb 21, 2013

- 739

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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- Admin
- #21

- Mar 5, 2012

- 8,899

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:

$$\lim_{x \to 0} \frac{\cos x - 1}{\sin x} = 0 \qquad \blacksquare$$

- Admin
- #22

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.

- Feb 13, 2012

- 1,704

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

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

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

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?...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

$\chi$ $\sigma$

- Thread starter
- #25

- Feb 21, 2013

- 739

Hello,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

$\chi$ $\sigma$

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?