# Homework Help: L'Hopitals Rule problem

1. Sep 24, 2012

### Bipolarity

1. The problem statement, all variables and given/known data

Evaluate the following limit

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

2. Relevant equations

3. 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

2. Sep 24, 2012

### mtayab1994

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. Sep 24, 2012

### Staff: Mentor

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. Sep 24, 2012

### Bipolarity

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

BiP

5. Sep 24, 2012

### Staff: Mentor

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. Sep 24, 2012

### SammyS

Staff Emeritus
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 $\displaystyle \cos(t)\text{csc}(t)\ .$