Finding the Limit of cos(pi*x / sinx)

  • Thread starter Oneiromancy
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In summary, the conversation was about finding the limit of a function using derivative rules or the difference quotient. One person suggested using the continuity of cosine for any real number, while another person mentioned using the quotient rule for finding the limit of sin(x)/x. The correct answer was found to be -1.
  • #1
Oneiromancy
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Find the limit.

limit x -> 0 cos(pi*x / sinx)

I'm only in cal I. I was hoping I wouldn't have to use the difference quotient so is there an easier way to do this using derivative rules or something?
 
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  • #2
[tex]lim_{x\rightarrow 0} cosf(x) =cos (lim_{x\rightarrow 0} f(x))[/tex]
 
  • #3
[tex]\lim_{x\rightarrow 0}\cos{\left(\frac{\pi x}{\sin x}\right)}[/tex]

Yes?
 
  • #4
rocophysics said:
[tex]\lim_{x\rightarrow 0}\cos{\left(\frac{\pi x}{\sin x}\right)}[/tex]

Yes?

Yeah this is what he meant, i guess. And follow rock.freak667's advice. But remember that this is true only because cos is continuous for any real.
[tex]lim_{x\rightarrow 0} cosf(x) =cos (lim_{x\rightarrow 0} f(x))[/tex]
Now can you see the trick after you go inside the cosine function with the limit?
 
  • #5
Right.

Can I just use the quotient rule on the inside?
 
  • #6
well you do not need to use the quotient rule, because do you know what the limit of

sin(x)/x is as x--->0

lim(x-->0)(sin(x))/x ----?
Just use this fact and you will be fine, because if you use the quotient rule as you are claimint to, you will get an intermediate form of 0/0.
 
  • #7
This would be x / sin x in this case. You got it backwards.
 
  • #8
Oneiromancy said:
This would be x / sin x in this case. You got it backwards.

NO, i did not get it backwards, because if you know what lim(x-->0)(sin(x))/x ----?
is you will not have problem finding out what
lim(x-->0)x/sinx is, since you will take its reciprocal and everything will turn into terms of sin(x)/x
 
  • #9
Oh, sorry, calm down lol. It's 1.
 
  • #10
[tex]\lim_{x\rightarrow 0}\cos{\left(\frac{\pi x}{\sin x}\right)}=\cos{(\lim_{x\rightarrow 0}\left\frac{\pi}{\frac{\sin x}{\ x}\right)}}[/tex]
do u see now what to do?
 
  • #11
Ya thanks it's -1.
 
  • #12
Oneiromancy said:
Ya thanks it's -1.

Good job!
 

Related to Finding the Limit of cos(pi*x / sinx)

What is the limit of cos(pi*x / sinx) as x approaches zero?

The limit of cos(pi*x / sinx) as x approaches zero is undefined, because the denominator approaches zero and the function becomes undefined.

How can we find the limit of cos(pi*x / sinx) as x approaches infinity?

To find the limit of cos(pi*x / sinx) as x approaches infinity, we can use the fact that cos(pi*x / sinx) will oscillate between -1 and 1 as x approaches infinity. Therefore, the limit does not exist.

What is the difference between a finite limit and an infinite limit?

A finite limit means that the function approaches a specific value as the input approaches a certain value. An infinite limit means that the function either approaches positive or negative infinity as the input approaches a certain value.

Can we use L'Hopital's rule to find the limit of cos(pi*x / sinx)?

Yes, we can use L'Hopital's rule to find the limit of cos(pi*x / sinx). We can rewrite the function as cos(pi*x) / sinx and take the derivative of both the numerator and denominator to simplify the expression.

Is the limit of cos(pi*x / sinx) equal to the limit of cosx / sinx as x approaches zero?

No, the limit of cos(pi*x / sinx) as x approaches zero is undefined, while the limit of cosx / sinx as x approaches zero is equal to 1. This is because the input in the first function is multiplied by pi, which changes the behavior of the function near zero.

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