How do you find the value of the limit as x approaches infinity of #sqrt(x^2+x+1)-x# by hand?

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Answer 1 · 2015-05-14T03:39:56

Turn it into a fraction, rationalize tlhe numerator, then find the limit of the fraction.

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

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

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

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

Recall: #sqrt(x^2)=absx#, but we want #lim_(xrarroo)#, so we're only interested in positive values of #x#. We'll replace #sqrt(x^2)# by #x#

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

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

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

Taking the limit as #xrarroo#, we get:

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