What is the limit of #sqrt(9x+x^2)/(x^4+7) # as x approaches infinity?

We have the limit

#lim_(xrarroo)sqrt(9x+x^2)/(x^4+7)#

Factor out the largest-degreed terms from the numerator and denominator of the fraction.

#=lim_(xrarroo)sqrt(x^2(9/x+1))/(x^4(1+7/x^4))#

Note that the #sqrt(x^2)# can be brought from the square root as just #x#.

#=lim_(xrarroo)(xsqrt(9/x+1))/(x^4(1+7/x^4))=lim_(xrarroo)(sqrt(9/x+1))/(x^3(1+7/x^4))#

When analyzing this as it goes to infinity, we see that #9/x# and #7/x^4# go to #0#.

#=sqrt(0+1)/(oo(0+1))=1/oo=0#

There is also a more intuitive approach to limits of this type.

In the numerator, we have in a square root a polynomial of degree #2#. Since there is a square root, the "overpowering" force, that is, the term that grows fast and determines most the growth of the function, is reduced to #sqrt(x^2)=x#, or a degree of #1#.

In the denominator, the overpowering term is of degree #4#.

Since #x^4# grows much faster than just #x#, and #x^4# is in the denominator, the limit will be #0#. Had the rational fraction been inverted, the limit as #xrarroo# would have been #oo#.