# Getting derivative of the following function using L'Hopital?

1. Dec 22, 2004

### KataKoniK

Getting limit of the following function using L'Hopital?

I have tried this question 5 times and each time I get a diff answer. I don't know what I am doing wrong, but here's the question. Any help?

lim
x-> 0

(4/x^2) - (2 / (1 - cos x))

edited - for typo in subject heading

Last edited: Dec 22, 2004
2. Dec 22, 2004

### HallsofIvy

Staff Emeritus
First, you don't mean "getting the derivative", you mean "finding the limit".

Second, exactly what is the function? The way you have written it, I would interpret it as (4/x^2)- 2/(1- cos x). Since the numerators of both fractions are non-zero while the denominators go to 0, there is NO limit. Certainly L'Hopital would not apply at all here.

3. Dec 22, 2004

### KataKoniK

Your interpretation is correct, so the answer DNE then? Furthermore, I just have a general question. How does 2sin x cosx−2x become sin 2x - 2x? I was just looking at past set solutions and it just suddenly does that. I don't understand

4. Dec 22, 2004

### Sirus

That is due to the following trig identity:
$$\sin{2x}=2\sin{x}\cos{x}$$

5. Dec 22, 2004

### KataKoniK

Got it. It's the double angle formula. Guess I just have to memorize it.

Last edited: Dec 22, 2004
6. Dec 22, 2004

### Sirus

That identity can be easily derived from the (hopefully) more familiar one below:
$$\sin{a+b}=\sin{a}\cos{b}+\sin{b}\cos{a}$$
Defining both $a$ and $b$ as the the same variable $x$, it is obvious that we get the previous identity for $\sin{2x}$. To be efficient at trigonometric limits, you should know these identities well, unless they are given to you in tests for reference.

7. Dec 23, 2004

### dextercioby

$$\lim_{x\rightarrow 0}\frac{4}{x^{2}}-\frac{2}{1-\cos x}=\lim_{x\rightarrow 0}\frac{4-4cosx-2x^{2}}{x^{2}(1-\cos x)}$$

=$$\lim_{x\rightarrow 0}\frac{4\sin x-4x}{2x(1-\cos x)+x^{2}\sin x}= 4\lim_{x\rightarrow 0}\frac{\cos x-1}{2(1-\cos x)+4x\sin x+x^{2}\cos x}$$

=$$-4\lim_{x\rightarrow 0}\frac{\sin x}{2\sin x+4\sin x+4x\cos x+2x\cos x-x^{2}\sin x}$$

=$$-4\lim_{x\rightarrow 0}\frac{\cos x}{6\cos x+6\cos x-8x\sin x-x^{2}\cos x}=-\frac{4}{12}=-\frac{1}{3}$$

Daniel.

Last edited: Dec 23, 2004
8. Dec 23, 2004

### dextercioby

Elegant way:
Use the fact that close to zero:
$$\cos x=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-...$$
,then the limit becomes:
$$\lim_{x\rightarrow 0}\frac{4}{x^{2}}-\frac{2}{1-1+\frac{x^{2}}{2}-\frac{x^{4}}{24}}=...=\lim_{x\rightarrow 0}\frac{48x^{2}-4x^{4}-48x^{2}}{12x^{4}-x^{6}}$$
=$$\lim_{x\rightarrow 0}\frac{-4}{12-x^{2}}=-\frac{1}{3}$$

Check it is still true for another term from the MacLaurin expansion of cosx
$$\cos x=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720}+...$$

Daniel.

9. Dec 23, 2004

### dextercioby

Check post number 7.I guess i applied it 4 times.

Daniel.

10. Dec 23, 2004

### KataKoniK

Thanks a lot Daniel. Saw my mistake!