Can L'Hopital's Rule Resolve This Trigonometric Limit?

[Bipolarity]

Homework Statement

Evaluate the following limit

$$\lim_{t→2π\mathbb{N}}\frac{sin(t)}{1-cos(t)}$$ for all natural numbers $\mathbb{N}$.

Homework Equations

The Attempt at a Solution

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

Am I correct?

BiP

 

[mtayab1994]

Homework Statement

Evaluate the following limit

$$\lim_{t→2π\mathbb{N}}\frac{sin(t)}{1-cos(t)}$$ for all natural numbers $\mathbb{N}$.

Homework Equations

The Attempt at a Solution

Plugging in $2π\mathbb{N}$ gives us 0 for both the numerator and denominator. Thus, we differentiate top and bottom to get $$\lim_{t→2π\mathbb{N}}\frac{cos(t)}{sin(t)}$$ which evaluates to $\frac{1}{0}$ 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.

 

[Mark44]

Homework Statement

Evaluate the following limit

$$\lim_{t→2π\mathbb{N}}\frac{sin(t)}{1-cos(t)}$$ for all natural numbers $\mathbb{N}$.

Homework Equations

The Attempt at a Solution

Plugging in $2π\mathbb{N}$ gives us 0 for both the numerator and denominator. Thus, we differentiate top and bottom to get $$\lim_{t→2π\mathbb{N}}\frac{cos(t)}{sin(t)}$$ which evaluates to $\frac{1}{0}$ 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?

 

[Bipolarity]

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

 

[Mark44]

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 -∞.

 

[SammyS]

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 $\displaystyle \frac{\cos(t)}{\sin(t)}$ as $\displaystyle \cos(t)\text{csc}(t)\ .$