# Homework Help: Calculus hw!

1. Sep 6, 2005

### zell_D

lim x->0 3sin4x/sin3x i do not know how to reduce the sin4x? and do i even use the property where lim x->0 sinx/x =1?

2. Sep 6, 2005

### LeonhardEuler

Do you know ll'Hospital's rule?

3. Sep 6, 2005

### zell_D

never heard of it, class just started so if i learned anything i forgot

4. Sep 6, 2005

### LeonhardEuler

If you have'nt done it, then don't use it. Use the formulas for the sine and cosine of double angles and sums of angles (i.e. use the sum of angles to make sin(3x)=sin(x+2x), into an expresion with only x and 2x, then use the double angle rules to make the functions of 2x into functions of x. Simmilarly, 4x=2(2x).)

5. Sep 6, 2005

### zell_D

ok thx another REAL DUMB question on my part lol havent done math for so long, when i make sin4x into sin2(2x) and then into 2sin2xcos2x, do i use distributive property with the 3?

6. Sep 6, 2005

### fourier jr

no, there's an identity that's probably on the inside cover of your textbook. it goes something like sin(a+b) = cos(a)sin(b) + cos(b)sin(a). i can't remember how that formula goes but it's something like that.

7. Sep 6, 2005

### LeonhardEuler

Yes, Fourier has it right. sin(3x)=sin(x+2x)=sin(x)cos(2x)+sin(2x)cos(x), then use the rule for sin(2x) again, and the rule cos(2x)=cos^2(x)-sin^2(x).

8. Sep 6, 2005

### zell_D

hmm can anyone check my work lol i feel dumb and still cant get the answer =/

lim x-> 0(from here on stated as lim)

lim 3sin4x/sin3x
= lim 3sin2(2x)/sin(x+2x)
= lim (3)2sin2xcos2x/sinxcos2x+cosxsin2x

what do i do after this?
do i need to use the other double angle property so the top looks like sin2xcos2x+cos2xsin2x?

9. Sep 6, 2005

### LeonhardEuler

No, the top will look like:
$$3\cdot2(2\sin{x}\cos{x}(\cos{x}^2-\sin{x}^2))$$

10. Sep 6, 2005

### zell_D

hmmm what can i cancel out?

Last edited: Sep 6, 2005
11. Sep 6, 2005

### LeonhardEuler

You just need to simplify it a little more with the double angle formulas in the denomenator to get a solution.

12. Sep 6, 2005

### zell_D

does it matter which form of cos' double angle formula i use?

13. Sep 6, 2005

### LeonhardEuler

No, it will take about the same amount of work whichever way.

14. Sep 6, 2005

### zell_D

lol... i ran outta room and still not finishing =/

do i use distributive property for the cosx^2-sinx^2?

ok i finished the answer is 4 is that correct? also i am still confused on one part, any confirmation would b nice

From:

lim 12sinxcosx(cosx^2-sinx^2)/3sinxcos^2(x)-sin^3(x)<-- can anyone check math on this part?
=lim 12sinxcos^3(x)-12sin^3(x)cosx/3sinxcos^2(x)-sin^3(x)
now is my problem, assuming i did i tall correctly, i know i factor sinx from the bottom, but how bout the 3?

Last edited: Sep 6, 2005
15. Sep 6, 2005

### lurflurf

The suggestions so far would work, but are not the best way.
$$\lim_{x\rightarrow 0}\frac{3\sin(4x)}{\sin(3x)}=\lim_{x\rightarrow 0}4\frac{\sin(4x)}{4x} \ \frac{3x}{\sin(3x)}=4\frac{L_1}{L_2}$$
Both of those limits are equal to the know limit for sin(x)/x
$$L_1=\lim_{x\rightarrow 0}\frac{sin(4x)}{4x}$$
$$L_2=\lim_{x\rightarrow 0}\frac{sin(3x)}{3x}$$
$$\lim_{x\rightarrow 0}\frac{sin(4x)}{4x}=\lim_{x\rightarrow 0}\frac{sin(3x)}{3x}=\lim_{x\rightarrow 0}\frac{sin(x)}{x}=1$$

Last edited: Sep 6, 2005
16. Sep 6, 2005

### LeonhardEuler

Not much further to go after you make those substitutions. Then you will be able to factor out a sin(x) that will leave some of the terms with only cos(x). Then you can take the limit, all the terms with sin(x) go to zero, all the terms with only cos(x) go to 1, and you have your answer.

17. Sep 6, 2005

### LeonhardEuler

Pretty clever.

18. Sep 6, 2005

### zell_D

care to explain? im sorta slow on math =/ this method is in my book i think but i do not get it lol

19. Sep 6, 2005

### LeonhardEuler

Ok, you know that
$$\lim_{x\rightarrow 0}\frac{\sin{x}}{x}=1$$
and that
$$\lim_{x\rightarrow 0}\frac{x}{\sin{x}}=1$$
Now suppose, for instance 4x=t:
$$\lim_{t\rightarrow 0}\frac{\sin{t}}{t}=1$$
$$=\lim_{4x\rightarrow 0}\frac{\sin{4x}}{4x}=1$$
So, in this problem what lurflurf did was to multiply the equation by $\frac{x}{x}$, which is always 1 as long as x is not 0, so it does not change the value. Then it becomes:
$$\lim_{x\rightarrow 0}\frac{3x\sin{4x}}{x\sin{3x}}$$
$$=\lim_{x\rightarrow 0}\frac{3\sin{4x}}{x}\frac{x}{\sin{3x}}$$
$$=\lim_{x\rightarrow 0}\frac{\sin{4x}}{x}\frac{3x}{\sin{3x}}$$
Now he multiplies by 1 as 4/4:
$$=\lim_{x\rightarrow 0}4\frac{\sin{4x}}{4x}\frac{3x}{\sin{3x}}$$
The limit of the product is the product of the limits, so:
$$=4\lim_{x\rightarrow 0}\frac{\sin{4x}}{4x}\lim_{x\rightarrow 0}\frac{3x}{\sin{3x}}$$