If a>0 and b>0 show that lim(sqrt((n+a)(n+b)) - n) = (a+b)/2?

Answer 1

[n→∞]lim √((n+a)(n+b)) - n

Multiply out the part in the square root:

[n→∞]lim √(n² + (a+b)n + ab) - n

Multiply both numerator and denominator by the conjugate:

[n→∞]lim ((n² + (a+b)n + ab) - n²)/(√(n² + (a+b)n + ab) + n)

Simplify:

[n→∞]lim ((a+b)n + ab)/(√(n² + (a+b)n + ab) + n)

Divide both numerator and denominator by n:

[n→∞]lim ((a+b) + ab/n)/(√(n² + (a+b)n + ab)/n + 1)

Simplify:

[n→∞]lim ((a+b) + ab/n)/(√(1 + (a+b)/n + ab/n²) + 1)

Now, evaluate the limit:

((a+b) + 0)/(√(1 + 0 + 0) + 1)
(a+b)/(√1 + 1)
(a+b)/2

And we are done. Note that the assumption that a>0 and b>0 was not required.

Answer 2

I assume that this should be the limit as n -> ∞.

lim (n->∞) √((n+a)(n+b)) - n
= lim (n->∞) n √((1+a/n) (1+b/n)) - n
= lim (n->∞) n √(1 + (a+b)/n + ab/n^2) - n
= lim (n->∞) n (1 + (a+b)/n + ab/n^2)^(1/2) - n
= lim (n->∞) n {1 + (1/2) [(a+b)/n + ab/n^2] + (1/2)(-1/2) [(a+b)/n + ab/n^2]^2 / 2! + ...} - n
= lim (n->∞) n {1 + (1/2)(a+b)/n + O(1/n^2)} - n
= lim (n->∞) (n + (1/2)(a+b) + O(1/n) - n)
= lim (n->∞) ((1/2)(a+b) + O(1/n))
= (1/2) (a+b).

Answer 3

um... i think you are wrong. it would equal (a+b)/2"<<3

this may be advance math for you because I am a physician mathematician

hope you get better answers than mine