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

Question

Evelyn Agard · Accepted Answer

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

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]}

Axel Alvarez · Answer

undefined To evaluate the limit of (x^4-10)/(4x^3+x) as x approaches infinity, we can use the concept of dominant terms. By dividing both the numerator and denominator by x^3, we get (x^4/x^3 - 10/x^3)/(4x^3/x^3 + x/x^3). Simplifying this expression, we have (x - 10/x^3)/(4 + 1/x^2). As x approaches infinity, the term 10/x^3 becomes negligible compared to x, and 1/x^2 becomes negligible compared to 4. Therefore, the limit simplifies to x/4. As x approaches infinity, the value of x/4 also approaches infinity. Hence, the limit of (x^4-10)/(4x^3+x) as x approaches infinity is infinity.