Question

How do you evaluate `[sin(3t) + 4t] / [t sec t]` as t approaches 0?

Answer 1

`lim_(t rarr 0) (sin(3t) + 4t) / (t sec(t)) = 7`

Explanation:

This is tricky.

Simplify `(sin(3t) + 4t) / (t sec(t))` into `(cos(t)*(sin(3t) + 4t)) / t`

Then apply some addition formulas and double-angle formulas.

Firstly let `sin(3t) = sin(2t + t)`
`= sin(2t)*cos(t) + cos(2t)*sin(t)`.

Then expand this with two compound angle formulas:

`sin(2t)*cos(t) + cos(2t)*sin(t)`
`= 2sin(t)*cos(t)*cos(t) + (2cos^2(t) - 1)*sin(t)`.
`= 2sin(t)*cos^2(t) +2cos^2(t)*sin(t) - sin(t)`
`= 4sin(t)*cos^2(t) - sin(t)`.

Then add the `4t`:
`4sin(t)*cos^2(t) - sin(t) + 4t`.

Then multiply by `cos(t)` so finally the original numerator is:
`4sin(t)*cos^3(t) - sin(t)*cos(t) + 4t*cos(t)`.

So, to find:
`lim_(t rarr 0) (4sin(t)*cos^3(t) - sin(t)*cos(t) + 4t*cos(t)) / t`

I shall steer clear of fancy formulas, and rely on an intuitive approach, if this suits.

Set the following values:
The approach of `t` to zero can be expressed as `t = 1/oo`.
Likewise, as `t` approaches zero, `sin(t) = 1/oo` and `cos(t) = 1`.

Hence, remove any `cos` terms from the expression.

So the numerator becomes:
`4sin(t)*cos^3(t) - sin(t)*cos(t) + 4t*cos(t)` `=4/oo - 1/oo + 4/oo`
`= 7/oo`

Finally, dividing by the `t` in the denominator:

`(7/oo)/ (1/oo) = 7`.