Question #d42b9

1 Answer
Mar 5, 2016

#lim_(xrarr-oo)(sqrt(x^2+3x)-x)=oo# and #lim_(xrarroo)(sqrt(x^2+3x)-x)=3/2#

Explanation:

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#