Limit involving sin at x=pi/6: Solving with Substitution Method

[mtayab1994]

Homework Statement

find the limit:

\lim_{x\rightarrow\frac{\pi}{6}}\frac{2sin(x)-1}{6x-\pi}

The Attempt at a Solution

Any starters please, I've tried the substitution method but went nowhere.I did x=(pi+h)/6. I still get 0 over something which is zero and i know that is not correct.

 

[jbunniii]

Can you please write out what you got when you tried the substitution?

 

[mtayab1994]

i got :

\lim_{x\rightarrow 0}\frac{2sin(\frac{\pi}{6}+\frac{h}{6})-1}{h}=\frac{0}{0}

and that's undefined.

 

[mtayab1994]

Any help please?

 

[Mark44]

This is an easy problem if you can use L'Hopital's Rule. I've tried several other approaches, but I have not hit on anything else that worked. That's not to say that there isn't another approach - I just wasn't able to stumble onto it.

 

[mtayab1994]

This is an easy problem if you can use L'Hopital's Rule. I've tried several other approaches, but I have not hit on anything else that worked. That's not to say that there isn't another approach - I just wasn't able to stumble onto it.

Yea with l'hopital's rule i come up with cos(x)/3 which then is easy to compute. But I have no clue on how to solve it without that.

 

[Mark44]

There probably is a way, but I have not been successful finding it. If you are allowed to use L'Hopital's Rule, I would use it.

 

[mtayab1994]

Yes, but too bad i can't use it. You've tried the substitution method and you keep getting 0/0 as well right?

 

[Dick]

Yea with l'hopital's rule i come up with cos(x)/3 which then is easy to compute. But I have no clue on how to solve it without that.

Try expressing the limit in terms of u=x-pi/6 as u->0. You should know the limits of sin(u)/u and (1-cos(u))/u.

 

[mtayab1994]

Yes, but that just keeps giving 0/0 which is undetermined.

 

[Dick]

Yes, but that just keeps giving 0/0 which is undetermined.

sin(u)/u is 0/0 but the limit as u->0 is 1. Just because it's 0/0 doesn't mean there is no limit.

 

[mtayab1994]

Yes i understand that part i know that lim x->0 sin(u)/U=1 but i don't know where I'm going to get the sin and the 1-cos(x) so i do (1-cos(x))/x^2=1/2. I have no clue how to reach that.

 

[Dick]

Yes i understand that part i know that lim x->0 sin(u)/U=1 but i don't know where I'm going to get the sin and the 1-cos(x) so i do (1-cos(x))/x^2=1/2. I have no clue how to reach that.

After the substitution sin(x) turns into sin(u+pi/6). Use the sin addition formula to break that up.

 

[mtayab1994]

ok i got:

\lim_{u\rightarrow0}\frac{2(sin(u)*cos(pi/6)+cos(u)*sin(pi/6))-1}{6u}

and when i sub in with 0 i got 0/0.

 

[Dick]

ok i got:

\lim_{u\rightarrow0}\frac{2(sin(u)*cos(pi/6)+cos(u)*sin(pi/6))-1}{6u}

and when i sub in with 0 i got 0/0.

Stop subbing in. Of course, it's still 0/0. What is sin(pi/6)? I do think you can express that in terms of limits of sin(u)/u and (1-cos(u))/u if you try.

 

[mtayab1994]

Of course i know sin(pi/6) is 1/2 but i don't get how you want me to express it.

 

[Dick]

Of course i know sin(pi/6) is 1/2 but i don't get how you want me to express it.

I would like you to express it as (something)*sin(u)/u+(something else)*(1-cos(u))/u.

 

[mtayab1994]

I understand what you mean 100% but I don't know where I'm going to get the 1-cos(x) from.

 

[Dick]

I understand what you mean 100% but I don't know where I'm going to get the 1-cos(x) from.

One of the terms in the numerator is 2*cos(u)*sin(pi/6), right?

 

[SammyS]

I understand what you mean 100% but I don't know where I'm going to get the 1-cos(x) from.

What's 2 times (1/2) ?

 

[mtayab1994]

What's 2 times (1/2) ?

that's 1

 

[mtayab1994]

One of the terms in the numerator is 2*cos(u)*sin(pi/6), right?

yes there is.

 

[Dick]

that's 1

Now we are getting someplace! :)

 

[mtayab1994]

Now we are getting someplace! :)

Ok and 1-1 in the numerator gives 0 doesn't it?

 

[Dick]

yes there is.

That is the cos(u) part.

 

[Dick]

Ok and 1-1 in the numerator gives 0 doesn't it?

NO. You get cos(u)-1. That's not zero.

 

[mtayab1994]

so 2*cos(u)*sin(pi/6) is equal to cos(u)?

 

[Dick]

so 2*cos(u)*sin(pi/6) is equal to cos(u)?

Why are you asking that?

 

[mtayab1994]

Why are you asking that?

because you said that 2*cos(u)*sin(pi/6) is the cos(u) part.

 

[mtayab1994]

Ok i got the limit as 1/(2√3) is that correct?

 

[Dick]

because you said that 2*cos(u)*sin(pi/6) is the cos(u) part.

Are you saying that you don't see why it is equal to cos(u)?

 

[mtayab1994]

Are you saying that you don't see why it is equal to cos(u)?

Is my answer of 1/(2√3) correct?

 

[Dick]

Is my answer of 1/(2√3) correct?

Yes, it is. That's the same thing l'Hopital would give you. You can always use l'Hopital to check these even if you can't use it to solve the problem.

 

[mtayab1994]

Yes, it is. That's the same thing l'Hopital would give you. You can always use l'Hopital to check these even if you can't use it to solve the problem.

Yes thank you, I really appreciate your help :) .