Evaluating Limits: How Do I Substitute Infinity Into the Equation?

In summary, the conversation discusses the evaluation of a limit involving 2πr and 2nRsin(π/n) as n approaches infinity. The solution involves canceling out the 2πr terms and using the fact that sin(x) is approximately equal to x for small values of x. The correct answer is 2Rπ.
  • #1
danago
Gold Member
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4
Hi. We havnt started limits in class yet, but I've come across one, which i need to evaluate.

[tex]
\mathop {\lim }\limits_{n \to \infty } 2\pi r + 2nR\sin \frac{\pi }{n}
[/tex]

For a limit like that, can i just substitute infinity into the equation? Or is there some rule against that? If so, would i be right in saying:

[tex]
\mathop {\lim }\limits_{n \to \infty } 2\pi r + 2nR\sin \frac{\pi }{n} = 2\pi r
[/tex]

Thanks for the help.
Dan.
 
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  • #2
Well sir dango

What your saying to me seams perfectly correct, however, limits are not my specialties. For the limit which you defined, that is the correct answer

unique_pavadrin

EDIT: dear sir dango what i had previously sated is incorrect, as n approaches n infinity, the function appraocd 2(pi)(r+R)
 
Last edited:
  • #3
[tex]
\mathop {\lim }\limits_{n \to \infty } 2\pi r + 2nR\sin \frac{\pi }{n} = 2\pi r
[/tex]

2pi r is a constant, take out of the limit. We have

[tex]
\mathop {\lim }\limits_{n \to \infty } 2nR\sin \frac{\pi }{n} = 2\pi r
[/tex]

Remember for small x, sin x=x? You can see that from the fact the tangent at x=0 is y=x (on the sine curve). So as n gets large, pi/n gets small.

The n's cancel so we get that limit is equal to 2Rpi, so the solution unique_pavadrin posted is correct.
 
  • #4
Gib Z said:
[tex]
\mathop {\lim }\limits_{n \to \infty } 2\pi r + 2nR\sin \frac{\pi }{n} = 2\pi r
[/tex]

2pi r is a constant, take out of the limit. We have

[tex]
\mathop {\lim }\limits_{n \to \infty } 2nR\sin \frac{\pi }{n} = 2\pi r
[/tex]

Remember for small x, sin x=x? You can see that from the fact the tangent at x=0 is y=x (on the sine curve). So as n gets large, pi/n gets small.

The n's cancel so we get that limit is equal to 2Rpi, so the solution unique_pavadrin posted is correct.

Ahh thanks very much for that. That answer is exactly what I am after, considering the context I am using the equation in :biggrin:
 
  • #5
Gib Z said:
[tex]
\mathop {\lim }\limits_{n \to \infty } 2\pi r + 2nR\sin \frac{\pi }{n} = 2\pi r
[/tex]

2pi r is a constant, take out of the limit. We have

[tex]
\mathop {\lim }\limits_{n \to \infty } 2nR\sin \frac{\pi }{n} = 2\pi r
[/tex]
No, you have cancel the 2 [itex]2\pir[/itex] terms:
[tex]
\mathop {\lim }\limits_{n \to \infty } 2nR\sin \frac{\pi }{n} = 0
[/tex][/quote]
Now what you say further makes sense.

Remember for small x, sin x=x? You can see that from the fact the tangent at x=0 is y=x (on the sine curve). So as n gets large, pi/n gets small.

The n's cancel so we get that limit is equal to 2Rpi, so the solution unique_pavadrin posted is correct.
 

1. What are the basic rules for evaluating limits?

The basic rules for evaluating limits are the sum rule, product rule, power rule, quotient rule, and constant multiple rule. The sum rule states that the limit of a sum is equal to the sum of the limits. The product rule states that the limit of a product is equal to the product of the limits. The power rule states that the limit of a power function is equal to the power of the limit. The quotient rule states that the limit of a quotient is equal to the quotient of the limits. The constant multiple rule states that the limit of a constant multiple of a function is equal to the constant times the limit of the function.

2. How do I use the sum and difference rules to evaluate limits?

To use the sum and difference rules, first determine if the limit is in the form of a sum or difference. If it is, you can apply the sum rule by taking the limit of each individual term and adding or subtracting them together. This is the same for the difference rule, except you would subtract the limits of each individual term instead of adding them.

3. Can the product rule be used to evaluate limits of two functions that are not continuous?

No, the product rule can only be used to evaluate limits when both functions are continuous at the point where the limit is being evaluated. If either function is not continuous at that point, the product rule cannot be applied.

4. What is the difference between the direct substitution and the limit laws methods for evaluating limits?

The direct substitution method involves simply substituting the value of the variable into the function and evaluating it. This method can only be used if the function is continuous at that point. The limit laws method involves using the basic rules for evaluating limits to simplify the function and then plugging in the value of the variable. This method can be used even if the function is not continuous at that point.

5. Can the L'Hopital's rule be used to evaluate limits of any function?

No, L'Hopital's rule can only be used to evaluate limits of functions that are in the form of an indeterminate form, such as 0/0 or ∞/∞. It is not applicable to limits of functions that are not in this form.

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