How do you find the limit of #1/(x^3 +4)# as x approaches #oo#?

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Answer 1 · 2017-02-19T00:53:57

# lim_(x rarr oo) 1/(x^3+4) = 0 #

We can multiply numerator and denominator by #1/x^3# (the reciprocal of the largest power in the denominator) as follows:

# lim_(x rarr oo) 1/(x^3+4) = lim_(x rarr oo) ((1/x^3)(1))/((1/x^3)(x^3+4))#
# " " = lim_(x rarr oo) (1/x^3)/(1+4/x^3)#

And we note that as #x rarr oo# then #1/x,1/x^2,1/x^3 rarr 0#, so:

# lim_(x rarr oo) 1/(x^3+4) = 0/(1+0)#
# " " = 0#

We can verify this result by looking at the graph of #y=1/(x^3+4)#

graph{ 1/(x^3+4) [-7, 13, -4.16, 5.84]}

and indeed it does appear that for large #x# the function is approaching a horizontal asymptote #y=0#