Trigonometry Limit of sinθ/θ without L'Hopital's Rule

[Robertoalva]

Homework Statement

Find the limit

Homework Equations

lim θ→ ∏/6 (sinθ - 1/2)/(θ - ∏/6)

The Attempt at a Solution

well, if you substitute the values, it gives 0/0, but that means that is undefined and that's not the answer. according to my book the answer is √(3)/2 and I am not supposed to use l'hospital's rule. this is we're I need help. Thanks

 

[Mark44]

Homework Statement

Find the limit

Homework Equations

lim θ→ ∏/6 (sinθ - 1/2)/(θ - ∏/6)

The Attempt at a Solution

well, if you substitute the values, it gives 0/0, but that means that is undefined and that's not the answer. according to my book the answer is √(3)/2 and I am not supposed to use l'hospital's rule. this is we're I need help. Thanks

Use a substitution (u = θ - $\pi$/6) to rewrite your limit as u → 0. Try to get it into the form $\lim_{u \to 0}\frac{sin(u)}{u}$, which is a well-known limit.

 

[lurflurf]

That is the derivative of sin(x) at x=∏/6. If you know the derivative of sine you are done. The derivative of sine can be expressed in terms of the derivative of sine when x=0. The derivative of sine at x=0 is usually an axiom or easily follows from the axioms.

 

[Robertoalva]

so basically the whole thing gives me 1? or the -1/2 stays and subtracts after i do the limit?

 

[Mark44]

so basically the whole thing gives me 1? or the -1/2 stays and subtracts after i do the limit?

Make the substitution and see what you get.

 

[Mark44]

That is the derivative of sin(x) at x=∏/6. If you know the derivative of sine you are done. The derivative of sine can be expressed in terms of the derivative of sine when x=0. The derivative of sine at x=0 is usually an axiom or easily follows from the axioms.

I'm guessing that the aim of the problem is to get the poster to evaluate the limit, but not in the context that it's the derivative of anything.

 

[Robertoalva]

what i have right now is lim u->0 (sin(u) - (1/2))/u. after that I know that lim u->0 sin(u)/u= 1 then 1-(1/2). but it gives me 1/2 and that's not the correct answer. what am I doing wrong?

 

[Robertoalva]

?

 

[SammyS]

what i have right now is lim u->0 (sin(u) - (1/2))/u. ...

That's not what you should be getting.

Show how you get that.

 

[Robertoalva]

that's how I did it... sin(u)/u= 1 so 1-(1/2)=1/2

 

[Mentallic]

1-(1/2)=1/2

What's the expression you ended up with to conclude this? Show us each step you've taken, starting with the u substitution.

 

[SammyS]

that's how I did it... sin(u)/u= 1 so 1-(1/2)=1/2

Mark 44 suggested that you use the substitution,

$\displaystyle u=\theta-(\pi/2) \$ .​

Of course, that gives

$\displaystyle \theta=u+(\pi/2)$
​

You need to use the angle addition formula for

$\displaystyle \sin\left(u+(\pi/2)\right)$​

 

[Robertoalva]

but he said that u= theta - (pi/6) not pi/2

 

[Robertoalva]

according to what i have read, sin(u+(pi/2))= cos u but, can I do it with sin(u+(pi/6)) ?

 

[Mentallic]

but he said that u= theta - (pi/6) not pi/2

He made a mistake.

according to what i have read, sin(u+(pi/2))= cos u but, can I do it with sin(u+(pi/6)) ?

No, you can't, but do you know how to expand sin(A+B)?

 

[Robertoalva]

yes sin(u+v)=sin(u)cos(v)+cos(u)sin(v) in which this case would be sin(u+(pi/2))= sin(u)cos(pi/2)+cos(u)sin(pi/2)

 

[SammyS]

but he said that u= theta - (pi/6) not pi/2

Yup, Mentallic is right.

I can't read my own posts to proof-read them.

 

[Mentallic]

yes sin(u+v)=sin(u)cos(v)+cos(u)sin(v) in which this case would be sin(u+(pi/2))= sin(u)cos(pi/2)+cos(u)sin(pi/2)

Ok, two problems. We're supposed to be making the substitution u = \theta - \pi/6 instead, and expressions like \cos(\pi/2) can be simplified.