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

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.

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) \to sqrt(1)=1#

As for the parenthesis in the denominator, with similar claims we have

#(1+cancel(7/x^4))\to 1#

Thus, the global ratio behaves like

#1/(infty*1)\to 0#

as #x# approaches positive infinity.