Question #77c61

1 Answer
Mar 2, 2017

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