How do you find the limit #(x+5)(1/(2x)+1/(x+2))# as #x->oo#?

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Answer 1 · 2016-11-18T13:41:56

First write it as a single ratio.

#(x+5)(1/(2x)+1/(x+2)) = (3x^2+17x+10)/(2x^2+4x)#

For #x != 0# we can factor out #x^2# in both the numerator and the denominator to get

# = (x^2(3+17/x+10/x^2))/(x^2(2+4/x))#

# = (3+17/x+10/x^2)/(2+4/x)# (for #x != 0#)

As #x# increases without bound all of #17/x#, #10/x^2# and #4/x# approach zero.

Therefore,

#lim_(xrarroo) (x+5)(1/(2x)+1/(x+2)) = lim_(xrarroo)(3+17/x+10/x^2)/(2+4/x)#

# = (3+0+0)/(2+0) = 3/2#