# Find the domain of definition and then the limit

1. Sep 14, 2006

### mohlam12

okay i have three limits, i did one and the two others i m stuck...
well here

1.
limit when x tends to zero of
(x-sin(px)) / (x-sin(qx))

p and q are positive integers.

for this one i have no idea what to do, i never worked with p or q...

2.
limit when x tends to zero of the function:
(x(1-cosx)) / (sin3x - 3sinx)
for this one, i expanded the denominator to get [ sinx(3cos²x-sin²x) ], then i finally have
(x(1-cosx)) / (sinx(3cos²x-sin²x))
and the limit is +infinty i think; am i right?

3.

f(x)=(x-1)/(x+1-|x|)
i first need to find the domain of definition, then find the limits in those points.

first, i need to separate that function into :

f(x)= (x-1)/(2x+1) [x<0] (x shouldn't equal 1/2)
f(x)= x+1 [x<o]

so, the limit in -infinity is -infinity
the limit in +infinity is 1/2
the limit in 1/2+ is -infinity
the limit in 1/2- is +infinity

is that right ^^
thanks!

Last edited: Sep 14, 2006
2. Sep 14, 2006

### VietDao29

Uhm... this may be a good candidate for L'Hopital's rule, or Taylor's expansion. Have you covered Taylor's expansion yet? Just expand it arround x = 0, and see if you get the answer.
Nope. :)
You should check it again:
sin(3x) = 3 sin(x) - 4 sin3(x), the denominator does not simplify to: sinx(3cos²x-sin²x)
x can be 1 / 2, the only value that makes f(x) undefined is x = - 1 / 2.
You should check the limits for x tends to -, and + infinity again, you seemed to have swapped the two.
Ok, can you go from here? :)

3. Sep 14, 2006

### daveb

L'Hospital's would be great for the 1st one, but since this is in Pre-Calc, I would assume you can't use anything with a deriviative yet.

I can't see what would be a good candidate, but maybe the Squeeze Theorem would work here.

4. Sep 14, 2006

### arildno

As for the first one, do the trick:
$$\frac{x-\sin(px)}{x-\sin(qx)}=\frac{x-px\frac{\sin(px)}{px}}{x-qx\frac{\sin(qx)}{qx}}=\frac{1-p\frac{\sin(px)}{px}}{1-q\frac{\sin(qx)}{qx}}$$

Last edited: Sep 14, 2006
5. Sep 14, 2006

### mohlam12

okay, for the second one, after simplifying i get -x/4sinx, and the limit of that when x tends to zero is -1/8. is this right?

the third one, (yes I meant x>o)

so, the limit in -infinity is 1/2
the limit in +infinity is 1/2
the limit in (-1/2)+ is -infinity
the limit in (-1/2)- is +infinity
is that right?

the first one,
I'd like to know how you went from $$\frac{x-\sin(px)}{x-\sin(qx)} to this \frac{x-px\frac{\sin(px)}{px}}{x-qx\frac{\sin(qx)}{qx}}$$

Thank you very much, I really appreciate your help :)

6. Sep 14, 2006

### arildno

Well, px/px=1, isn't it?
We can multiply any term we wish with 1, since 1 times a number equals that number..

7. Sep 14, 2006

### VietDao29

This one is still wrong. :) Re-check it the third time to see why.

8. Sep 14, 2006

### mohlam12

Got it thanks!

Oooops!! it's plus infinity... I forgot that there was an absolute value!! Thank you =]

9. Sep 15, 2006

### mohlam12

Okay I have a problem now...

for the second one,

(x(1-cosx)) / (sin3x - 3sinx)

How did you get sin(3x) = 3 sin(x) - 4 sin^3 (x), I didn't get that. and let's suppose it is right... now my function is:

-x/4sinx(1+cosx)

i dont know what do afterwards ?

EDIT: NEVER MIND! THANKS

Last edited: Sep 15, 2006