How do you evaluate the limit #(sqrt(x+1)-2)/(x-3)# as x approaches #3#?

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Answer 1 · 2016-08-04T14:00:44

#= 1/4#

#lim_(x to 3) (sqrt{(x+1)}-2)/(x-3)#

with #x = 3 + h, h = x-3, 0 < abs(h) "<<" 1#

becomes #lim_(h to 0) (sqrt{3 + h +1}-2)/(3+ h-3)#

#lim_(h to 0) 1/h * (sqrt{4 + h}-2)#

#lim_(h to 0) 1/h *( 2 sqrt{1 + h/4}-2)#

by Taylor expansion

#lim_(h to 0) 1/h * ( 2 (1 + 1/2 * h/4 + mathcal(O)(h^2)) -2 )#

#lim_(h to 0) 1/h * 2 ( 1/2 * h/4 + mathcal(O)(h^2)) #

#lim_(h to 0) 1/h * ( h/4 + mathcal(O)(h^2)) #

#lim_(h to 0) 1/4 + mathcal(O)(h) = 1/4#

You can see clearly that the limit is the same as x approaches 3 on both sides as the sign of h is in the error term.