How do you evaluate the limit #(x^4+3^x)/(x^5+1)# as x approaches #oo#?

Question

Emilia Aguilar · Accepted Answer

#oo# #lim_(x to oo) (x^4+3^x)/(x^5+1)# this is in #oo/oo# indeterminate form and therefore L'Hopital's rule applies on repeated basis as follows #= lim_(x to oo) (4x^3+ln (3) 3^x)/(5x^4)# #= lim_(x to oo) (12x^2+ln^2 (3) 3^x)/(20x^3)# #= lim_(x to oo) (24x+ln^3 (3) 3^x)/(60x^2)# #= lim_(x to oo) (24+ln^4 (3) 3^x)/(120x)# #= lim_(x to oo) (ln^5 (3) 3^x)/(120)# #= infty#

Scarlett Allison · Answer

undefined To evaluate the limit as x approaches infinity of (x^4+3^x)/(x^5+1), we can use the concept of dominant terms.

As x approaches infinity, the term with the highest power in the numerator and denominator will dominate the expression. In this case, the dominant terms are x^4 in the numerator and x^5 in the denominator.

Dividing both the numerator and denominator by x^5, we get (x^4/x^5 + 3^x/x^5)/(x^5/x^5 + 1/x^5).

Simplifying further, we have (1/x + (3^x)/(x^5))/(1 + 1/x^5).

As x approaches infinity, 1/x approaches 0, and (3^x)/(x^5) also approaches 0. Therefore, the expression simplifies to (0 + 0)/(1 + 0), which is equal to 0.

Hence, the limit of (x^4+3^x)/(x^5+1) as x approaches infinity is 0.