# Trigonometric Limit Problem

1. Apr 23, 2015

### terryds

1. The problem statement, all variables and given/known data
Find the limit of :
lim x-> (π/2) (2-2sin x)/(6x-3π)

2. The attempt at a solution

lim x-> (π/2) (2-2sin x)/(6x-3π)
=lim x-> (π/2) 2-2 sin x / 6 (x- (1/2)pi)

Assuming that y = x - (π/2)
So,
lim y->0 (2-2sin(y+pi/2))/6y
lim y->0 (2-2 (sin y cos pi/2 + cos y sin pi/2)/6y

Then, I substitute y=0 into the sin y cos pi/2 , so the equation remains :

lim y-> 0 (2-2 cos y sin pi/2)/6y
lim y->0 2(1-cos y sin pi/2) /6y
lim y-> 0 2 (2sin^2((1/2)y) sin pi/2) /6y
lim y->0 (4 sin^2 ((1/2)y) sin(pi/2)) /6y
lim y->0 (2/3) (sin^2 (1/2)y) (2*(1/2)y) (sin(pi/2))

Then, I'm stuck here, because I need one more (1/2)y as the denominator of sin(1/2)y , because sin^2(1/2)y = sin(1/2)y * sin(1/2)y, so I need two (1/2) y as the denominator
But, I can just make one. If I make two and normalize it, it'll be zero.. But, I'm not quite sure..

2. Apr 23, 2015

### paisiello2

Maybe try L'Hopital's rule?

3. Apr 23, 2015

### SammyS

Staff Emeritus
Maybe not -- in the pre-calculus forum .

4. Apr 23, 2015

### Staff: Mentor

Try to keep in mind the mathematical level of the student who posts a problem. Help for a problem posted in the Precalc section should generally not use ideas or concepts from calculus. Occasionally a member will post a problem in the wrong homework section, but we mentors make an effort to move such posts to the right places.

5. Apr 23, 2015

### SammyS

Staff Emeritus
No need to substitute y = 0 into sin(y) there, because cos(π/2) =0 so that sin(y)cos(π/2) =0 for all y ..

Also, there is no need for you to drag around sin(π/2) because, sin(π/2) = 1.

6. Apr 23, 2015

### paisiello2

Aren't limits a calculus concept?

7. Apr 23, 2015

### SammyS

Staff Emeritus
Sometimes yes, sometimes no. They are usually introduced before differentiation, so L'Hôpital's rule wouldn't be appropriate in that case.

8. Apr 23, 2015

### Staff: Mentor

You're right Sammy. My focus was on thinking that this problem was one in which $\lim_{x \to 0}\frac{1 - cos(x)}{x}$ was supposed to be used, and didn't notice the effect on the original problem caused by a changed sign. I've deleted that post of mine.

9. Apr 23, 2015

### SammyS

Staff Emeritus
me too -- deleted it.

It still comes to something of the form $\displaystyle \ \lim_{y \to 0}\frac{1 - \cos(y)}{y} \$ .

Last edited: Apr 23, 2015
10. Apr 24, 2015

### terryds

Hmm.. You're right..
Then, what should I do after :
lim y->0 (2/3) (sin(pi/2)) (sin^2 (1/2)y)/(2*(1/2)y)
??

If I make it like below
lim y->0 (2/3) (sin(pi/2)) (sin^2(1/2)y) / ((4y)((1/2)y)^2)
The limit will be undetermined (division by zero)
Then, what should I do ?

11. Apr 24, 2015

### SammyS

Staff Emeritus
( You continue to keep sin(π/2) in your expressions even though sin(π/2) = 1 . )

Have you learned that $\displaystyle\ \lim_{x\to 0}\frac{\sin(x)}{x}=1 \$ ?

12. Apr 24, 2015

### terryds

Yes, I have learned that..
That's why I make the equation :
lim y->0 (2/3) (sin(pi/2)) (sin^2(1/2)y) / ((4y)((1/2)y)^2)

I notice the sin^2(1/2)y/((1/2)y)^2 becomes 1
But, there is 4y in the denominator (to normalize the equation).. And if I plugged y = 0, it will be a division by zero

13. Apr 24, 2015

### SammyS

Staff Emeritus
That would make the limit undefined. Right ?

BUT, that 'extra' y should be in the numerator, not in the denominator.

14. Apr 24, 2015

### terryds

Yeah, you're right.. Got it! The limit is zero