Question

How do you find the limit of `(sin^2(4t)) / t^2` as t approaches 0?

Answer 1

Use algebra, properties of limits and the important limit `lim_(xrarr0)sinx/x=1`

Explanation:

`sin^2(4t)/t^2 = (sin(4t)/t)^2` (Algebra)

So we need only find `lim_(trarr0)sin(4t)/t` and square that result. (Property of limits)

There are 3 big tricks in mathematics:

1. Add `0`
2. Multiply by `1`
3. Change the problem into something we know how
   to solve

In this case we'll use item 2. to accomplish 3.

`sin(4t)/t = 4/4 sin(4t)/t = 4(sin(4t)/((4t)))`

Now, as `trarr0`, we also see that `4trarr0`, so thinking of `4t` as `x` in `lim_(xrarr0)sinx/x=1`,

we arrive at `lim_(trarr0)(sin(4t)/((4t)))=1`

So `lim_(trarr0)4(sin(4t)/((4t))) = 4(1) = 4`

And finally

`(lim_(trarr0)sin(4t)/t)^2 = (lim_(trarr0)4(sin(4t)/4t))^2`

`= (4)^2 = 16`