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

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In summary, the conversation revolves around finding the limit of a function using different methods. The person who asked the question has tried using the substitution method and L'Hopital's rule but was unable to find a solution. Others suggest using a different approach and eventually the person is able to find the limit as 1/(2√3). The conversation also touches on the concept of 0/0 being an indeterminate form and how it can still have a limit.
  • #1
mtayab1994
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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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  • #2
Can you please write out what you got when you tried the substitution?
 
  • #3
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
Any help please?
 
  • #5
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
Mark44 said:
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
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
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
mtayab1994 said:
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
Yes, but that just keeps giving 0/0 which is undetermined.
 
  • #11
mtayab1994 said:
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
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
mtayab1994 said:
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
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
mtayab1994 said:
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
Of course i know sin(pi/6) is 1/2 but i don't get how you want me to express it.
 
  • #17
mtayab1994 said:
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
I understand what you mean 100% but I don't know where I'm going to get the 1-cos(x) from.
 
  • #19
mtayab1994 said:
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
mtayab1994 said:
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) ?
 
  • #21
SammyS said:
What's 2 times (1/2) ?

that's 1
 
  • #22
Dick said:
One of the terms in the numerator is 2*cos(u)*sin(pi/6), right?

yes there is.
 
  • #23
mtayab1994 said:
that's 1

Now we are getting someplace! :)
 
  • #24
Dick said:
Now we are getting someplace! :)

Ok and 1-1 in the numerator gives 0 doesn't it?
 
  • #25
mtayab1994 said:
yes there is.

That is the cos(u) part.
 
  • #26
mtayab1994 said:
Ok and 1-1 in the numerator gives 0 doesn't it?

NO. You get cos(u)-1. That's not zero.
 
  • #27
so 2*cos(u)*sin(pi/6) is equal to cos(u)?
 
  • #28
mtayab1994 said:
so 2*cos(u)*sin(pi/6) is equal to cos(u)?

Why are you asking that?
 
  • #29
Dick said:
Why are you asking that?

because you said that 2*cos(u)*sin(pi/6) is the cos(u) part.
 
  • #30
Ok i got the limit as 1/(2√3) is that correct?
 
  • #31
mtayab1994 said:
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)?
 
  • #32
Dick said:
Are you saying that you don't see why it is equal to cos(u)?

Is my answer of 1/(2√3) correct?
 
  • #33
mtayab1994 said:
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.
 
  • #34
Dick said:
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 :) .
 

1. What is the substitution method for solving a limit involving sin at x=pi/6?

The substitution method involves replacing the variable in the limit expression with a new variable that approaches the desired value. In this case, we would replace x with a new variable, let's say u, and then take the limit as u approaches pi/6.

2. Why is the substitution method used for solving limits involving sin at x=pi/6?

The substitution method is used because it allows us to simplify the expression and evaluate the limit more easily. In this case, replacing x with u will allow us to use the trigonometric identity sin(pi/6) = 1/2, which will make the limit easier to evaluate.

3. What is the first step in solving a limit involving sin at x=pi/6 using the substitution method?

The first step is to replace the variable x with a new variable, let's say u, in the limit expression. Then, we take the limit as u approaches pi/6.

4. Can the substitution method be used for solving limits involving other trigonometric functions?

Yes, the substitution method can be used for solving limits involving any trigonometric function. The key is to choose the appropriate substitution that will simplify the expression and make it easier to evaluate the limit.

5. Are there any limitations to using the substitution method for solving limits involving sin at x=pi/6?

Yes, there are some limitations to using the substitution method. It may not always work for more complex limit expressions, and in some cases, it may not be the most efficient method for solving the limit. It is important to consider other methods as well when solving limits involving trigonometric functions.

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