Question #77c61

1 Answer
Eddie
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#

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