How do you prove that the limit of #sqrt(x+1) = 2 # as x approaches 3 using the epsilon delta proof?

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Answer 1 · 2017-02-22T15:15:33

See explanation below

Choose any #epsilon > 0# and evaluate the difference:

#abs (sqrt(x+1) -2)#

we have:

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

#abs (sqrt(x+1) -2) = abs ( (x+1) -4 )/ (sqrt(x+1) +2)#

#abs (sqrt(x+1) -2) = abs ( x-3)/ (sqrt(x+1) +2)#

Now consider that:

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

is a strictly decreasing function, so for #x in (3-delta, 3+delta)#:

#1/(sqrt(x+1) +2) <= 1/(sqrt(3-delta+1) +2)= 1/(sqrt(4-delta) +2)#

and clearly:

#abs (x- 3) <= delta#

so:

#abs (sqrt(x+1) -2) <= delta/(sqrt(4-delta) +2)#

So, if we choose #delta_epsilon# such that:

#delta_epsilon/(sqrt(4-delta_epsilon) +2) < epsilon#

we have:

#abs (x-3) < delta_epsilon => abs (sqrt(x+1) -2) < epsilon#