How do you determine the limit of #sqrt( x^2 + x) / (-2x)# as n approaches #-oo#?

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Answer 1 · 2016-04-12T12:12:34

The limit is #1/2#. For "how" please see the explanation below.

Here are the crucial observations:

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

(The above is true provided that #x !=0#. But when evaluating limits as #x# increases or decreases without bound, we aren't concerned with #x=0#.)

Also observe/recall that for real #x#, we have

#sqrt(x^2) = absx = {(-x,"if",x<0),(x,"if",0 <= x) :}#.

Prepared with these two observations, we proceed:

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

# = lim_(xrarr-oo)((-x)sqrt(1+1/x))/(-2x)# #" "# [we have #x < 0#]

# = lim_(xrarr-oo)(sqrt(1+1/x))/(2)#

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

Bonus
Here is the limit as #x# increases without bound. (I'm not a fan of the expression "approaches #oo#".)

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

# = lim_(xrarroo)((x)sqrt(1+1/x))/(-2x)# #" "# [we have #x > 0#]

# = lim_(xrarroo)(sqrt(1+1/x))/(-2)#

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