How do you find the limit of #sqrt(9x+x^2)/(x^4+7)# as x approaches #oo#?

Question

Christopher Agresta · Accepted Answer

Reqd. Lim.#=0#.

Reqd. Limit #=lim_(xrarroo)(sqrt(9x+x^2))/(x^4+7)# #=lim_(xrarroo)(sqrt(x^2(9/x+1)))/(x^4(1+7/x^4)# #=lim_(xrarroo)(xsqrt(9/x+1))/(x^4(1+7/x^4))# #=lim_(xrarroo)(1/x^3)(sqrt(9/x+1))/(1+7/x^4)# We need to recall, here, that, as #xrarroo, 9/x=9*1/xrarr0, 1/x^3rarr0, and, 7/x^4rarr0#. Hence, the reqd. lim.#=0*(sqrt(0+1)/(1+0))=0*1=0#. In fact, #lim _(xrarroo)sqrt(9/x+1)# #=sqrt{lim_(xrarroo)(9/x+1)# #=sqrt(0+1)=1#. The inter-changeability of the limit & sqrt. fun is because of the continuity of the sqrt. fun.

Scarlett Allison · Answer

The limit exists, and it is zero.

Factor out the greatest power of #x# from both numerator and denominator: #sqrt(9x+x^2)/(x^4+7) = sqrt(x^2(9/x+1))/(x^4(1+7/x^4)# Since #sqrt(x^2)=|x|#, we can continue with #sqrt(x^2(9/x+1))/(x^4(1+7/x^4)) = (|x|sqrt(9/x+1))/(x^4(1+7/x^4)# Since #x# is approaching positive infinity, we have #|x|=x#. The next step is thus #(|x|sqrt(9/x+1))/(x^4(1+7/x^4)) = sqrt(9/x+1)/(x^3(1+7/x^4))# At this point, we're good to go: since #x# is approaching positive infinity, every quantity like #k/x^alpha# vanished, with #k# a real number and #alpha>0#. Thus, the square root approaches one: #sqrt(cancel(9/x)+1) o sqrt(1)=1# As for the parenthesis in the denominator, with similar claims we have #(1+cancel(7/x^4)) o 1# Thus, the global ratio behaves like #1/(infty*1) o 0# as #x# approaches positive infinity.

Emily Ammerman · Answer

undefined To find the limit of sqrt(9x+x^2)/(x^4+7) as x approaches infinity, we can use the concept of limits.

First, we divide both the numerator and denominator by x^2 to simplify the expression:

sqrt((9/x) + 1)/(x^2/x^2 + 7/x^2)

Simplifying further, we get:

sqrt((9/x) + 1)/(1 + 7/x^2)

As x approaches infinity, 9/x and 7/x^2 both approach zero. Therefore, the expression simplifies to:

sqrt(0 + 1)/(1 + 0)

Which is equal to:

sqrt(1)/1

And the square root of 1 is equal to 1. Therefore, the limit of sqrt(9x+x^2)/(x^4+7) as x approaches infinity is 1.