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lim as x-->a of 1/[sqrt(x+1)] = 1/[sqrt(a+1)] for all a > -1.

Please show your work. Thank you!

2007-02-14 21:35:11 · 1 answers · asked by Anonymous in Science & Mathematics Mathematics

1 answers

for proofing that u have to do this:
for all eps>0 there is a delta>0 s.t.
0<|x-a| |1/(x+1)^1/2-1/(a+1)^1/2| |1/(x+1)^1/2-1/(a+1)^1/2|=
|((x+1)^1/2-(a+1)^1/2))/((x+1)(a+1))^1/2)|=
|(x-a)/(((x+1)(a+1))^1/2)*(x+1)^1/2 +(a+1)^1/2))|
< (since |x-a| eps

2007-02-14 22:01:11 · answer #1 · answered by vinchenzo_corleone 2 · 0 0

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