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

1 Answer
Andrea S.
Jan 9, 2017

#lim_(x->oo) (x^4-10)/(4x^3+x) = +oo#

Explanation:

When evaluating the limit of a rational function for #x->+-oo# you can ignore all the monomials above and below the line except the ones with highest order.

So:

#lim_(x->oo) (x^4-10)/(4x^3+x) = lim_(x->oo) x^4/(4x^3) = lim_(x->oo) x/4 = +oo#

You can see that by separating the sum:

#(x^4-10)/(4x^3+x) = x^4/(4x^3+x) -10/(4x^3+x) = 1/((4x^3+x)/x^4) -10/(4x^3+x) = 1/(4/x+1/x^3) -10/(4x^3+x) #

Evidently:

#lim_(x->oo) 10/(4x^3+x) = 0#

#lim_(x->oo) 1/(4/x+1/x^3) = +oo#

graph{(x^4-10)/(4x^3+x) [-10, 10, -5, 5]}

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