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
