Question

Lim 1 - cos 40x>01 - cos 60​

Answer 1

Answer:

The answer is 4/9 if the problem is:

$\lim_{\theta \rightarrow 0}\frac{1-\cos(4\theta)}{1-\cos(6\theta)}$.

Step-by-step explanation:

I think this says:

$\lim_{\theta \rightarrow 0}\frac{1-\cos(4\theta)}{1-\cos(6\theta)}$.

Please correct me if I'm wrong about the problem.

Here are some useful limits we might use:

$\lim_{u \rightarrow 0}\frac{\sin(u)}{u}=1$

$\limg_{u \rightarrow 0}\frac{\cos(u)-1}{u}=0$

So for our limit... I'm going to multiply top and bottom by the conjugate of the bottom; that is I'm going to multiply top and bottom by $1+\cos(6\theta)$:

$\lim_{\theta \rightarrow 0}\frac{1-\cos(4\theta)}{1-\cos(6\theta)}\cdot\frac{1+\cos(6\theta)}{1+\cos(6\theta)}$

When you multiply conjugates you only have to do first and last of FOIL:

$\lim_{\theta \rightarrow 0}\frac{(1-\cos(4\theta))(1+\cos(6\theta))}{1-\cos^2(6\theta)}$

By the Pythagorean Identities, the denominator is equal to $\sin^2(6\theta)$:

$\lim_{\theta \rightarrow 0}\frac{(1-\cos(4\theta))(1+\cos(6\theta))}{\sin^2(6\theta)}$

I'm going to divide top and bottom by $36\theta^2$ in hopes to use the useful limits I mentioned:

$\lim_{\theta \rightarrow 0}\frac{\frac{(1-\cos(4\theta))(1+\cos(6\theta))}{36\theta^2}}{\frac{\sin^2(6\theta)}{36\theta^2}}$

Let's tweak our useful limits I mentioned so it is more clear what I'm going to do in the following steps:

$\lim_{\theta \rightarrow 0}\frac{\sin(6\theta)}{6\theta}=1$

$\lim_{\theta \rightarrow 0}\frac{\cos(4\theta)-1}{4\theta}=0$

The bottom goes to 1.  The limit will go to whatever the top equals if the top limit exists.  

So let's look at the top in hopes it goes to a number:

$\lim_{\theta \rightarrow 0}\frac{1-\cos(4\theta)}{36\theta^2} \cdot (1+\cos(6\theta)}$

We are going to multiple the first factor by the conjugate of the top; that is we are multiply top and bottom by $1+\cos(4\theta)$:

$\lim_{\theta \rightarrow 0}\frac{1-\cos(4\theta)}{36\theta^2} \cdot \frac{1+\cos(4\theta)}{1+\cos(4\theta)} \cdot (1+\cos(6\theta)}$

Recall the thing I said about multiplying conjugates:

$\lim_{\theta \rightarrow 0}\frac{1-\cos^2(4\theta)}{36\theta^2} \cdot \frac{1+\cos(6\theta)}{1+\cos(4\theta)}$

We are going to apply the Pythagorean Identities here:

$\lim_{\theta \rightarrow 0}\frac{\sin^2(4\theta)}{36\theta^2} \cdot \frac{1+\cos(6\theta)}{1+\cos(4\theta)}$

$\lim_{\theta \rightarrow 0}\frac{\sin^2(4\theta)}{\frac{9}{4}(4\theta)^2} \cdot \frac{1+\cos(6\theta)}{1+\cos(4\theta)}$

$\lim_{\theta \rightarrow 0}\frac{4}{9}\frac{\sin^2(4\theta)}{(4\theta)^2} \cdot \frac{1+\cos(6\theta)}{1+\cos(4\theta)}$

Ok this looks good, we are going to apply the useful limits I mentioned along with substitution to find the remaining limits:

$\frac{4}{9}(1)^2 \frac{1+\cos(6(0))}{1+\cos(4(0))}$

$\frac{4}{9}(1)\frac{1+1}{1+1}$

$\frac{4}{9}(1)\frac{2}{2}$

$\frac{4}{9}(1)$

$\frac{4}{9}$

The limit is 4/9.