How do you evaluate the limit #(3x^4-x^2+5)/(10-2x^4)# as x approaches #oo#?

1 Answer
Steve M
Dec 15, 2016

# lim_(x rarr oo) (3x^4-x^2+5)/(10-2x^4) = -3/2#

Explanation:

Note that as #x rarr oo# then #1/x,1/x^2,1/x^3 rarr 0#

We can therefore multiply numerator and denominator by #1/x^4# (the reciprocal of the largest power in the denominator) as follows:

# lim_(x rarr oo) (3x^4-x^2+5)/(10-2x^4) = lim_(x rarr oo) (3x^4-x^2+5)/(10-2x^4)*(1/x^4)/(1/x^4)#
# " " = lim_(x rarr oo) ((1/x^4)(3x^4-x^2+5))/((1/x^4)(10-2x^4))#
# " " = lim_(x rarr oo) (3-1/x^2+5/x^4)/(10/x^4-2)#
# " " = (3-0+0)/(0-2)#
# " " = -3/2#

We can verify this result by looking at the graph of #y=(3x^4-x^2+5)/(10-2x^4)#
graph{(3x^4-x^2+5)/(10-2x^4) [-10, 10, -5, 5]}
and indeed it does appear that for large #x# the function is approaching a horizontal asymptote #y=-3/2#

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