Prove the statement using the epsilon-delta definition of a limit

Question

lim [root(x)+1] / x+1

x->1

Answer 1

Definition: 0 < | x - 1| < delta ----> | f(x) - 1 | < epsilon

-epsilon < root(x) - 1 < epsilon

1 - epsilon < root(x) < 1 + epsilon

(1- epsilon)^2 < x < (1+epsilon)^2

1 - 2 * epsilon + epsilon^2 < x < 1 + 2*epsilon + epsilon^2

epsilon^2 - 2 * epsilon + 1 < x < epsilon^2 + 2*epsilon + 1

Meanwhile

-delta < x-1 < delta

1 - delta < x < 1+delta

Let delta =

Min {| 1 - (epsilon^2 - 2*epsilon + 1) |, | 1 - (epsilon^2 + 2 * epsilon + 1)| }

=Min { | 1 - epsilon^2 + 2*epsilon - 1 | , | 1 - epsilon^2 - 2*epsilon - 1| }

=Min { | epsilon^2 - 2*epsilon|, | epsilon^2 + 2 * epsilon | }

= | epsilon^2 - 2*epsilon|

0 < |x - 1 | < | epsilon^2 - 2*epsilon|

-|epsilon^2 - 2*epsilon| < x-1 < |epsilon^2 - 2*epsilon| < |epsilon^2 + 2*epsilon|

= epsilon^2 + 2*epsilon

Note that for x>0 and y>0, |x-y| = |x + -y| <= |x| + |-y| = |x| + |y|

Multiplying by -1, =|x-y| >= -(|x|-|y|)

the chained inequality becomes:

-(|epsilon^2| + |2*epsilon|)< -|epsilon^2 - 2*epsilon| < x-1 < |epsilon^2 - 2*epsilon|

< |epsilon^2 + 2*epsilon|

= epsilon^2 + 2*epsilon

- (epsilon^2 + 2*epsilon) - (-1) < x < epsilon^2 + 2 * epsilon + 1

- (epsilon^2 + 2*epsilon - 1) < x < epsilon^2 + 2 * epsilon + 1

0 < x < (epsilon+1)^2

0 < sqrt(x) < epsilon + 1

0 < sqrt(x) 1 < epsilon

This proves sqrt(x) tends to 1 and x tends to 1

LImit of x and x tends to 1 is proven to be 1

simply by letting delta equal epsilon

combining these limits concludes the proof, since adding 1 to

each limit does not affect the proof.

Answer 2

This is too lengthy an answer to type out, but I can help you with problems of this type. You need to show that for any eps > 0, there exists a del > 0 such that if 0 < |x-1| < del, |f(x)-1| < eps, where f(x) = [root(x)+1] / x+1. (The limit is 1 by substitution, though your question does not show it, unless you have a typo in the problem.)