# Homework Help: Limits involving sin.

1. Feb 27, 2012

### mtayab1994

1. The problem statement, all variables and given/known data

find the limit:

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

3. 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.

Last edited: Feb 27, 2012
2. Feb 27, 2012

### jbunniii

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

3. Feb 27, 2012

### mtayab1994

i got :

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

and that's undefined.

Last edited by a moderator: Feb 27, 2012
4. Feb 27, 2012

### mtayab1994

5. Feb 27, 2012

### Staff: 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.

6. Feb 27, 2012

### mtayab1994

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.

7. Feb 27, 2012

### Staff: 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.

8. Feb 27, 2012

### 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?

9. Feb 27, 2012

### Dick

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.

10. Feb 27, 2012

### mtayab1994

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

11. Feb 27, 2012

### Dick

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.

12. Feb 27, 2012

### 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.

13. Feb 27, 2012

### Dick

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

14. Feb 27, 2012

### 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.

15. Feb 27, 2012

### Dick

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.

16. Feb 27, 2012

### mtayab1994

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

17. Feb 27, 2012

### Dick

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

18. Feb 27, 2012

### mtayab1994

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

19. Feb 27, 2012

### Dick

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

20. Feb 27, 2012

### SammyS

Staff Emeritus
What's 2 times (1/2) ?

21. Feb 27, 2012

### mtayab1994

that's 1

22. Feb 27, 2012

### mtayab1994

yes there is.

23. Feb 27, 2012

### Dick

Now we are getting someplace! :)

24. Feb 27, 2012

### mtayab1994

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

25. Feb 27, 2012

### Dick

That is the cos(u) part.