Question

Evaluate `lim _(x-> oo) (sinxcosx)/(3x)` ?

Answer 1

`lim _(xrarroo) (sinxcosx)/(3x) =0`

Explanation:

Use the Squeeze Theorem at infinity.

Since, `sinx` and `cosx` are between `-1` and `1`, so is their product.

`-1 <= sinxcosx <= 1`

We are interested in `xrarroo`, so we are interested in positive values of `x`.
When `x` is positive, `3x` is positive, so we can divide the inequality without reversing the directions of the inequalities.

`(-1)/(3x) <= (sinxcosx)/(3x) <= 1/(3x)`

`lim_(xrarroo)(-1)/(3x) = 0` and `lim_(xrarroo)(1)/(3x) = 0`.

Therefore, `lim _(xrarroo) (sinxcosx)/(3x) =0`.