Question

Question #77c61

Answer 1

`oo`

Explanation:

`lim_(x to oo) x (3+sinx)`

`= 3 lim_(x to oo) x + lim_(x to oo) x sinx`

Because of the periodic nature of the second term, with `alpha = sin x` we can see it as:

`implies 3 lim_(x to oo) x + alpha lim_(x to oo) x` where `alpha in [-1,1]`

`= (3 + alpha) lim_(x to oo) x`

And because `(3 + alpha) in [2,4]`, the limit is `oo`