L'Hopitals Rule problem

  • Thread starter Bipolarity
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In summary: If t is close to k(2##\pi##), then \cos(t)\text{csc}(t)\approx 1, and so the limit is ∞. If t is farther away from k(2##\pi##), then \cos(t)\text{csc}(t)\approx -1, and so the limit is -∞.
  • #1
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Homework Statement



Evaluate the following limit

[tex] \lim_{t→2π\mathbb{N}}\frac{sin(t)}{1-cos(t)} [/tex] for all natural numbers [itex]\mathbb{N}[/itex].

Homework Equations


The Attempt at a Solution


Plugging in [itex] 2π\mathbb{N} [/itex] gives us 0 for both the numerator and denominator. Thus, we differentiate top and bottom to get [tex] \lim_{t→2π\mathbb{N}}\frac{cos(t)}{sin(t)} [/tex] which evaluates to [itex]\frac{1}{0}[/itex] which does not exist.

Am I correct?

BiP
 
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  • #2
Bipolarity said:

Homework Statement



Evaluate the following limit

[tex] \lim_{t→2π\mathbb{N}}\frac{sin(t)}{1-cos(t)} [/tex] for all natural numbers [itex]\mathbb{N}[/itex].

Homework Equations





The Attempt at a Solution


Plugging in [itex] 2π\mathbb{N} [/itex] gives us 0 for both the numerator and denominator. Thus, we differentiate top and bottom to get [tex] \lim_{t→2π\mathbb{N}}\frac{cos(t)}{sin(t)} [/tex] which evaluates to [itex]\frac{1}{0}[/itex] which does not exist.

Am I correct?

BiP

Use l'hopital's rule the second time to get -sin(t)/cos(t) and evaluate that to see what you get. But "1/0" is the correct way of writing 1/0 and that is infinity if you're dealing with limits.
 
  • #3
Bipolarity said:

Homework Statement



Evaluate the following limit

[tex] \lim_{t→2π\mathbb{N}}\frac{sin(t)}{1-cos(t)} [/tex] for all natural numbers [itex]\mathbb{N}[/itex].

Homework Equations





The Attempt at a Solution


Plugging in [itex] 2π\mathbb{N} [/itex] gives us 0 for both the numerator and denominator. Thus, we differentiate top and bottom to get [tex] \lim_{t→2π\mathbb{N}}\frac{cos(t)}{sin(t)} [/tex] which evaluates to [itex]\frac{1}{0}[/itex] which does not exist.

Am I correct?

BiP

You probably should elaborate a bit. Here are two examples of limits where the denominator approaches 0.

$$ \text{1}~\lim_{x \to 0} \frac{1}{x}$$
$$ \text{2}~\lim_{x \to 0} \frac{1}{x^2}$$

For the first, the limit does not exist in any sense, since the left- and right-side limits are different. For the second, the limit also does not exist, in the sense of being a finite number, but we say that the limit is ∞.

For your problem, is the limit ∞ or -∞, or does it fail to exist at all?
 
  • #4
Mark44 said:
You probably should elaborate a bit. Here are two examples of limits where the denominator approaches 0.

$$ \text{1}~\lim_{x \to 0} \frac{1}{x}$$
$$ \text{2}~\lim_{x \to 0} \frac{1}{x^2}$$

For the first, the limit does not exist in any sense, since the left- and right-side limits are different. For the second, the limit also does not exist, in the sense of being a finite number, but we say that the limit is ∞.

For your problem, is the limit ∞ or -∞, or does it fail to exist at all?

How can I show which one of these is the case in my problem?

BiP
 
  • #5
Bipolarity said:
How can I show which one of these is the case in my problem?

BiP
In the expression cos(t)/sin(t), when t is close to k(2##\pi##), the numerator is close to 1, and the denominator is close to 0. The key is to determine what the denominator is doing when t is also close to this value. If the denominator changes sign, then the limit is undefined. If the denominator doesn't change sign, then the limit is either ∞ or -∞.
 
  • #6
mtayab1994 said:
Use l'hopital's rule the second time to get -sin(t)/cos(t) and evaluate that to see what you get. But "1/0" is the correct way of writing 1/0 and that is infinity if you're dealing with limits.
You can't use L'Hôpital's rule a second time. after on application, you don't have an indeterminate form.

Bipolarity,

Look at [itex]\displaystyle \frac{\cos(t)}{\sin(t)}[/itex] as [itex]\displaystyle \cos(t)\text{csc}(t)\ .[/itex]
 

1. What is L'Hopital's Rule?

L'Hopital's Rule is a mathematical principle that can be used to evaluate limits of indeterminate forms, such as 0/0 or ∞/∞. It states that if the limit of a fraction is in an indeterminate form, then the limit of the quotient of the derivatives of the numerator and denominator is equal to the original limit.

2. When is L'Hopital's Rule useful?

L'Hopital's Rule is useful when evaluating limits that involve fractions with a variable in the denominator, where direct substitution results in an indeterminate form. It can also be used to simplify complex limits and to prove the convergence or divergence of a series.

3. How do you apply L'Hopital's Rule?

To apply L'Hopital's Rule, you must first identify whether the limit is in an indeterminate form. If it is, then take the derivative of the numerator and denominator separately. If the resulting limit is still in an indeterminate form, repeat the process until a definitive value is obtained or the limit is shown to be infinite.

4. Are there any limitations to L'Hopital's Rule?

Yes, L'Hopital's Rule can only be applied to limits involving fractions with a variable in the denominator. It cannot be used for limits that do not have a fraction, and it cannot be used to evaluate limits involving trigonometric functions or exponentials.

5. Can L'Hopital's Rule be used for infinite limits?

Yes, L'Hopital's Rule can be used for infinite limits as long as the limit is in an indeterminate form. In this case, the resulting limit may be a finite value, or it may show that the original limit is infinite.

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