Can you evaluate this limit

L = lim [b-b√(1-2k/b)] as b→∞

It's an indeterminate of form 'infinity minus infinity'.
Hint: The answer should be L = k.

I am assuming that we need to calculate:
lim (b-->infinity) [b - b√(1 - 2k/b)].

By multiplying b - b√(1 - 2k/b) by [b + b√(1 - 2k/b)]/[b + b√(1 - 2k/b)], we have:
lim (b-->infinity) [b - b√(1 - 2k/b)]
= lim (b-->infinity) {[b - b√(1 - 2k/b)][b + b√(1 - 2k/b)]}/[b + b√(1 - 2k/b)]
= lim (b-->infinity) [b^2 - b^2(1 - 2k/b)]/[b + b√(1 - 2k/b)], via difference of squares
= lim (b-->infinity) (b^2 - b^2 + 2kb)/[b + b√(1 - 2k/b)]
= 2k * lim (b-->infinity) b/[b + b√(1 - 2k/b)]
= 2k * lim (b-->infinity) 1/[1 + √(1 - 2k/b)], by dividing num/denom by b
= (2k)[1/(1 + 1)]
= k, as required.

I hope this helps!

L = lim [b-b√(1-2k/b)] as b→∞

(1) b-b√(1-2k/b) = b[1 - √(1-2k/b)] = b[1 - (1-2k/b)^(1/2)]

(2) 1 - (1-2k/b)^(1/2) = 1 - 1^(1/2) - (- 1/2) (2k/b) + (-2k/b) terms of increasing order

In the limit, those (-2k/b) terms of increasing order converge to 0 as b→∞; so

lim b[1 - (1-2k/b)^(1/2)] = lim (1/2)(2k) = k = L

Go to this website and click "show steps:"

http://www.wolframalpha.com/input/?i=lim…