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L'Hopitals Rule problem

  • Thread starter Bipolarity
  • Start date
  • #1
775
1

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
 

Answers and Replies

  • #2
584
0

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
33,514
5,195

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
775
1
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
33,514
5,195
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
SammyS
Staff Emeritus
Science Advisor
Homework Helper
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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]
 

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