How do you find the limit of #[4x+sinx] / [2x-sinx]# as x approaches infinity?

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Answer 1 · 2018-04-05T09:47:46

Divide with #x# denominator and numerator to get

#(4+sinx/x)/(2-sinx/x)#

Because the following holds

#|sinx|<=1#

Hence

#|sinx|/x<=1/x#

As #x->oo# we have that #sinx/x->0#

Hence the requested limit is

#lim_(x->oo)(4+sinx/x)/(2-sin/x)=(4+0)/(2-0)=2#