The definition of the limit is
#lim_(x->x_0)f(x)=L#
#AA epsilon>0, EE delta>0 # such that
#|f(x)-L|< epsilon#
when #|x- x_0| < delta#
Here,
#f(x)=1/sqrt(x-2)#
#x_0=3#
#L=1#
#epsilon=0.01#
Therefore,
#|1/sqrt(x-2)-1|<0.01#
#-0.01<1/sqrt(x-2)-1<0.01#
#-0.01+1<1/sqrt(x-2)<0.01+1#
#0.99<1/sqrt(x-2)< 1.01#
#1/0.99>sqrt(x-2)>1/1.01#
#1.02>(x-2)>0.98#
#3.02>x>2.98#
#2.98 < x < 3.02#
Also,
#|x-3| < delta#
#-delta<x3 < delta#
#3-delta< x < 3+delta#
Therefore,
#3-delta=2.98#
#delta=3-2.98=0.02#
#3+delta=3.02#
#delta=3.02-3=0.02#
I hope that this will help
Addition
#|1/sqrt(x-2)-1|<epsilon#
#-epsilon<1/sqrt(x-2)-1<epsilon#
#1-epsilon<1/sqrt(x-2)<1+epsilon#
#1/(1-epsilon)>sqrt(x-2)>1/(1+epsilon)#
#1/(1-epsilon)^2>(x-2)>1/(1+epsilon)^2#
#2+1/(1-epsilon)^2>x>2+1/(1+epsilon)^2#
Therefore,
#2+1/(1+epsilon)^2< x <2+1/(1-epsilon)^2#
#2+1/(1+epsilon)^2-3< x -3<2+1/(1-epsilon)^2-3#
#1/(1+epsilon)^2-1< x -3<1/(1-epsilon)^2-1#
But, #3-delta< x <3+delta#
#-delta< x-3 <delta#
#-delta=1/(1+epsilon)^2-1#
#delta=1/(1-epsilon)^2-1#
As #epsilon# is very small, the 2 values of #delta# are equivalent
