How do you find #lim (x-1)/(x-2sqrtx+1)# as #x->1^+#?
1 Answer
Mar 2, 2017
See below.
Explanation:
# = (sqrtx+1)/(sqrtx-1)# #" "# for#x != 1#
# = 2/0^+ = oo#
The notation in the last line is intended to indicate that the numerator approaches
The result is a quotient that is increasing without bound.