In evaluating limits at infinity, we are not interested in what happens when #x=0#, so we can write
#sqrt(x^2+3x)=sqrt(x^2(1+3/x)) = sqrt(x^2)sqrt(1+3/x)#
Recall that #sqrt(x^2) = absx#, so we may proceed:
#lim_(xrarr-oo)(sqrt(x^2+3x)-x)=lim_(xrarr-oo)(sqrt(x^2)sqrt(1+3/x)-x)#
# = lim_(xrarr-oo)(-xsqrt(1+3/x)-x)#
# = lim_(xrarr-oo)(-x(sqrt(1+3/x)+1))#
# = -(-oo)(2) = oo#
(This result can be seen more quickly, if less formally, by observing that ar #xrarr-oo#, both #sqrt(x^2+3x)# and #-x# increase without bound, so their sum increases without bound. That is, the initial form of the limit is #oo-(-oo)# which gives a limit of #oo#.)
#lim_(xrarroo)(sqrt(x^2+3x)-x)# has initial form #oo-oo# which is indeterminate.
#(sqrt(x^2+3x)-x) = (sqrt(x^2+3x)-x)/1 * (sqrt(x^2+3x)+x)/(sqrt(x^2+3x)+x) #
# = ((x^2+3x)-x^2)/(sqrt(x^2+3x)+x)#
# = (3x)/(sqrt(x^2)sqrt(1+3/x)+x)#
# = (3x)/(xsqrt(1+3/x)+x)# #" "# (for #x > 0#, we have #sqrt(x^2) = x#)
# = 3/(sqrt(1+3/x)+1)#
#lim_(xrarroo)3/(sqrt(1+3/x)+1) = 3/2#
