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What is the limit of #(sqrt(x^2+x)-x) # as x approaches infinity?

1 Answer
Feb 17, 2016

Answer:

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

Explanation:

The initial form for the limit is indeterminate #oo-oo#

So, use the conjugate.

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

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

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

#lim_(xrarroo) x/(sqrt(x^2+x)+x)# has indeterminate form #oo/oo#, but we can factor and reduce.

We know that #sqrt(x^2)=absx#, so for positive #x# (which is all we are concerned about for a limit as #x# increases without bound) we have

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

# = x/(xsqrt(1+1/x)+x)# #" "# (for # x > 0#)

# = x/(x(sqrt(1+1/x)+1))#

# = 1/(sqrt(1+1/x)+1)#

#lim_(xrarroo)1/(sqrt(1+1/x)+1) = 1/(sqrt1+1)=1/2#