use formal definition of a limit to prove: lim sqrt x =1 when x approaches to 1

Question

Use epsilon-delta language to prove the limit

Answer 1

0<|x-1|<d⇒|sqrt(x)-1|<e, where d represents delta and e represents epsilon. This statement is equivalent to limit as x approaches 1 of sqrt(x)=1; for all d satisfying the first inequality we need to prove there is e satisfying the second inequality. The first inequality can be written:

0<|(sqrt(x)-1)(sqrt(x)+1)|<d⇒0<|sqrt(x)-1|(sqrt(x)+1)<d. Substituting the second inequality: 0<e(sqrt(x)+1)<d, so e<d/(sqrt(x)+1), implying that e is less than d, and where x=1, e<d/2. The factor sqrt(x)+1 is always positive, with a minimum value of 1 and a maximum value of 2, so the inequality remains valid and there is always a value of e for every value of d.