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Limits involving sin.

  • Thread starter mtayab1994
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  • #1
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Homework Statement



find the limit:

[tex]\lim_{x\rightarrow\frac{\pi}{6}}\frac{2sin(x)-1}{6x-\pi}[/tex]

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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Answers and Replies

  • #2
jbunniii
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Can you please write out what you got when you tried the substitution?
 
  • #3
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i got :

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

and that's undefined.
 
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  • #4
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Any help please?
 
  • #5
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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
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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.
 
  • #7
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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
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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
Dick
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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.
 
  • #10
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Yes, but that just keeps giving 0/0 which is undetermined.
 
  • #11
Dick
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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.
 
  • #12
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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
Dick
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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.
 
  • #14
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ok i got:

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

and when i sub in with 0 i got 0/0.
 
  • #15
Dick
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ok i got:

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

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.
 
  • #16
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Of course i know sin(pi/6) is 1/2 but i don't get how you want me to express it.
 
  • #17
Dick
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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.
 
  • #18
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I understand what you mean 100% but I don't know where i'm going to get the 1-cos(x) from.
 
  • #19
Dick
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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?
 
  • #20
SammyS
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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) ?
 
  • #22
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One of the terms in the numerator is 2*cos(u)*sin(pi/6), right?
yes there is.
 
  • #23
Dick
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  • #24
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Now we are getting someplace! :)
Ok and 1-1 in the numerator gives 0 doesn't it?
 
  • #25
Dick
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