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

1 Answer
Jim H · mason m
Apr 26, 2016

#lim_(xrarroo)x/sqrt(x^2-x) = 1# Bonus: #lim_(xrarr-oo)x/sqrt(x^2-x) = -1#

Explanation:

Note that, for all #x# other than #0# (which we are not interested for limits at infinity), we have

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

Note also that #sqrt(x^2) = absx = {(x,"if", x >= 0),(-x,"if",x < 0):}#

So we get

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

# = lim_(xrarroo)x/(xsqrt(1-1/x)# #" "# (As #xrarroo# we are only interested in positive values of #x#.)

# = lim_(xrarroo)1/(sqrt(1-1/x)) = 1/sqrt(1-0) = 1#

Bonus

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

# = lim_(xrarr-oo)x/(-xsqrt(1-1/x)# #" "# (As #xrarr-oo# we are only interested in negative values of #x#.)

# = lim_(xrarroo)1/(-sqrt(1-1/x)) = 1/(-sqrt(1-0)) = -1#

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