1. ## Calculas- limits

Apply limit laws to evaluate or show that the limit does not exist.

lim Cos (π + ∆ x) +1
∆x -> 0 ∆x

I started like this but It didnt go anywhere:

lim Cos (π + ∆ x) +1 . Cos (π + ∆ x) -1 = lim Cos (π + ∆ x)^2 -1
∆x -> 0 ∆x Cos (π + ∆ x) -1 ∆x -> 0 ∆x(Cos(π + ∆ x) -1

2. Hi, meme22!

Originally Posted by meme22
Apply limit laws to evaluate or show that the limit does not exist.

lim Cos (π + ∆ x) +1
∆x -> 0 ∆x

I started like this but It didnt go anywhere:

lim Cos (π + ∆ x) +1 . Cos (π + ∆ x) -1 = lim Cos (π + ∆ x)^2 -1
∆x -> 0 ∆x Cos (π + ∆ x) -1 ∆x -> 0 ∆x(Cos(π + ∆ x) -1
I assume this is the problem you are trying to solve:

$\text{Evaluate }\lim_{\Delta x\to0}\frac{\cos\left(\pi + \Delta x\right) + 1}{\Delta x}\text{.}$

Is that right?

First of all, you can easily see what this limit should be if you observe that

$\lim_{\Delta x\to0}\frac{\cos\left(\pi + \Delta x\right) + 1}{\Delta x} = \lim_{\Delta x\to0}\frac{\cos\left(\pi + \Delta x\right) - (-1)}{\Delta x}$

$= \lim_{\Delta x\to0}\frac{\cos\left(\pi + \Delta x\right) - \cos\left(\pi\right)}{\Delta x} = \cos'\left(\pi\right) = -\sin\left(\pi\right) = 0$.

So, we should suspect the limit to be zero. Now, to solve it using the properties of limits, here are some hints:

1. Use the fact that $\cos\left(u\pm v\right) = \cos u\cos v\mp\sin u\sin v$ to rewrite the numerator

2. Make sure that you are familiar with this special limit:

$\lim_{x\to0}\frac{\cos\left(x\right) - 1}x = 0$

Does that help?

3. Yes, thank you I figured it out