# Limits involving sin.

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.

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Homework Helper
Gold Member
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.

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mtayab1994

Mentor
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.

Mentor
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?

Homework Helper
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.

Homework Helper
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.

Homework Helper
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.

Homework Helper
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.

Homework Helper
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.

Homework Helper
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?

Staff Emeritus
Homework Helper
Gold Member
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.

Homework Helper
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?

Homework Helper
yes there is.

That is the cos(u) part.

Homework Helper
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)?

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

mtayab1994

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?

Homework Helper
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?

Homework Helper
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 :) .