How do you find the limit of # [ x + sqrt (x^2 + 2x) ]# as x approaches #-oo#?

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Answer 1 · 2016-08-15T09:02:32

#= -1#

#lim_(x to - oo) [ x + sqrt (x^2 + 2x) ]#

prepare for binomial expansion
#lim_(x to - oo) [ x + |x| sqrt (1 + 2/x) ]#

expand out
#lim_(x to - oo) [ x + |x| (1 + 1/2 2/x + O(1/x^2)) ]#

#lim_(x to - oo) [ x + |x| + |x|/x + |x| O(1/x^2) ]#

for #x <0#

#lim_(x to - oo) [ x + (-x) + (-x)/x + (-x) O(1/x^2) ]#

#lim_(x to - oo) [ -1 -x O(1/x^2) ]#

#lim_(x to - oo) [ -1 + O(1/x) ]#

#= -1#

Answer 2 · 2016-09-20T20:54:37

To find this limit without binomial expansion see below.

#lim_(xrarr-oo)(x+sqrt(x^2+2x))# has indeterminate form #-oo+oo#

Rewrite as a fraction and rationalize the numerator.

#((x+sqrt(x^2+2x)))/1* ((x-sqrt(x^2+2x)))/((x-sqrt(x^2+2x)))#

# = (x^2-(x^2+2x))/(x-sqrt(x^2+2x))#

# = (-2x)/(x-sqrt(x^2(1+2/x))# #" "# for #x != 0#

# = (-2x)/(x-sqrt(x^2)sqrt(1+2/x))#

Recall that #sqrt(x^2) = absx# so for #x < 0# we have #sqrt(x^2) = -x#.

Our expression is equivalent to

# = (-2x)/(x+xsqrt(1+2/x))# #" "# for #x < 0 #

# = (-2x)/(x(1+sqrt(1+2/x)))# #" "# for #x < 0 #

# = (-2)/(1+sqrt(1+2/x))# #" "# for #x < 0 #

Evaluating the limit as #xrarr-oo# yields

#(-2)/(1+sqrt(1+0)) = (-2)/2 = -1#