Question

How do you find `\lim _ { \theta \rightarrow 0} \frac { \cos ( 7\theta ) - 1} { \sin ( 4\theta ) }`?

Answer 1

`lim_(theta->0)(cos(7theta)-1)/sin(4theta)=0`

Explanation:

We first just try plugging in `0` to see what we get:
`(cos(7*0)-1)/sin(4*0)=(1-1)/0=0/0`

This results in the indeterminate form `0/0`, which doesn't help us.

There is however a trick you can use to evaluate limits of this indeterminate form. It's called L'Hôpital's Rule. It basically says that if both the top and the bottom tend to `0`, we can take the derivative of both the top and bottom, knowing that the limit will be the same. Formally, this is:
If `lim_(x->c) f(x)=0` and `lim_(x->c) g(x)=0` and `lim_(x->c) (f'(x))/(g'(x))=K`
Then `lim_(x->c)f(x)/g(x)=K`

Applying this to our case, we get:
`lim_(theta->0)(cos(7theta)-1)/sin(4theta)=lim_(theta->0)(d/(d\theta)(cos(7theta)-1))/(d/(d\theta)(sin(4theta))`

The derivatives can be computed using the chain rule:
`d/(d\theta)(cos(7theta)-1)=-7sin(7theta)`

`d/(d\theta)(sin(4theta))=4cos(4theta)`

Now we put back into the limit and evaluate at `0`:
`lim_(theta->0)(-7sin(7theta))/(4cos(4theta))=(-7sin(0))/(4cos(0))=(-7*0)/(4*1)=0/4=0`