Question

How do you find the limit of `tan(4alpha)/sin(2alpha)` as `alpha->0`?

Answer 1

`2`

Explanation:

`lim_(alpha to 0) (tan4alpha)/(sin 2alpha)`

This is in `0/0` indeterminate form so L'Hopital is a possibility.

We do it algebraically, using:

`sin 2 z = 2 sin z cos z`, and

`cos 2z = cos^2 z - sin^2 z` and variations of this and the Pythagorean ID.

We can say:

`= lim_(alpha to 0) (sin 4alpha)/(cos 4 alpha) 1/(sin 2alpha)`

`= lim_(alpha to 0) (2 sin 2alpha cos 2 alpha)/(2 cos^2 2alpha - 1) 1/(sin 2alpha)`

`= lim_(alpha to 0) (2 cos 2 alpha)/(2 cos^2 2alpha - 1)`

These are both continuous at `alpha = 0` so we can plug in `alpha = 0` to get:

`= lim_(alpha to 0) (2 * 1)/(2 *1 - 1) = 2`

Answer 2

To solve without using derivatives (l'Hospital's Rule) see below.

Explanation:

`tan(4alpha) = sin(4alpha)/cos(4alpha)`

Use the sine of a double angle formula to rewrite this

`= (2sin(2alpha)cos(2alpha))/cos(4alpha)`

So

`tan(4alpha)/sin(2alpha) = (2sin(2alpha)cos(2alpha))/(sin(2alpha)cos(4alpha))`

`= (2cos(2alpha))/(cos(4alpha))`

Evaluating the limit as `alpha rarr0` we get

`(2cos(0))/cos(0) = (2(1))/(1) = 2`

Answer 3

`lim_(alphararr0)tan(4alpha)/sin(2alpha)=2`

Explanation:

We can also use the standard limit `lim_(xrarr0)sin(x)/x=1`. This can be inverted to say that `lim_(xrarr0)x/sin(x)=1`.

We can rearrange the given limit to make these appear:

`lim_(alphararr0)tan(4alpha)/sin(2alpha)=lim_(alphararr0)sin(4alpha)1/cos(4alpha)1/sin(2alpha)`

Multiplying by `alpha/alpha`:

`=lim_(alphararr0)sin(4alpha)/alpha1/cos(4alpha)alpha/sin(2alpha)`

Multiplying by `(2(2))/4`:

`=2(lim_(alphararr0)sin(4alpha)/(4alpha)1/cos(4alpha)(2alpha)/sin(2alpha))`

We can split up the limit. We also see that `lim_(alphararr0)sin(4alpha)/(4alpha)` and `lim_(alphararr0)(2alpha)/sin(2alpha)` are in the form `lim_(xrarr0)sin(x)/x=1`.

`=2*lim_(alphararr0)(2alpha)/sin(2alpha)*lim_(alphararr0)1/cos(4alpha)*lim_(alphararr0)(2alpha)/sin(2alpha)`

`=2*1*1/cos(0)*1`

`=2`