# Limits - Formula Validation/Verification

1. Jan 21, 2016

### YoshiMoshi

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

The picture attached appeared in my powerpoint for my class. It's been a long time since I took calculus 1, but if I remember correctly this formula is wrong correct?

I mean thinking about it
limit k-> inf ( cos(theta)^k ) = 0 if theta is not a multiple of pi OR +/- 1 if theta is a multiple of pi

limit k-> inf ( sin(theta)^k ) = 0 if theta is not a multiple of pi/2 OR +/- 1 if theta is a multiple of pi/2

I don't see how the generic formula is correct since domain of theta was not provided.

Thanks in advance.

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### Capture.PNG
File size:
956 bytes
Views:
41
2. Jan 21, 2016

### Ray Vickson

You are more-or-less on the right track, but some of your details need tightening up. First, it is enough to assume $0 \leq \theta < 2 \pi$, because $\sin$ and $\cos$ are periodic, with period $2 \pi$.

Looking at $\sin \theta$, the limit $\lim_{k \to \infty} \sin^k \theta = 0$ whenever $|\sin \theta | < 1$, so whenever $\theta \neq \pi/2, 3\pi/2.$ At $\theta = \pi/2$ we have $\sin(\pi/2) = 1$, so $\sin^k (\pi/2) \to 1$ as $k \to \infty$. However, $\sin(3\pi/2) = -1$, so $\sin^k (3 \pi/2) = (-1)^k$ does not have a limit as $k \to \infty$. It is 100% wrong to say that the limit is $\pm 1$ in this case: it isn't, there is no limit!.

You can do the same type of thing for $\cos \theta$.

BTW: to avoid trouble, you need to assume that $k \to \infty$ through integer values, because if $k$ is just some large real number, using it as a power could involve taking fractional powers of negative numbers, and that can get tricky.

Last edited: Jan 21, 2016
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted