Trigonometric Limit Problem

[terryds]

Homework Statement

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..
Please help me..

 

[paisiello2]

Maybe try L'Hopital's rule?

 

[SammyS]

Maybe try L'Hopital's rule?

Maybe not -- in the pre-calculus forum .

 

[Mark44]

Maybe try L'Hopital's rule?

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.

 

[SammyS]

Homework Statement

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)
...

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.

 

[paisiello2]

Try to keep in mind the mathematical level of the student who posts a problem...

Aren't limits a calculus concept?

 

[SammyS]

Aren't limits a calculus concept?

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

 

[Mark44]

Actually, Mark, your limit looks to be unbounded. That in the OP has the form 0/0 .

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.

 

[SammyS]

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.

me too -- deleted it.

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

 

[terryds]

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.

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 ?

 

[SammyS]

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 ?

( 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 \$ ?

 

[terryds]

( 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 \$ ?

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

 

[SammyS]

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

That would make the limit undefined. Right ?

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

 

[terryds]

That would make the limit undefined. Right ?

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

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