Prove that the limit function is 1/x?

Question

Prove that as k -> 0, the following expression reduces to 1/x:

(1/(k(k-√2))) ( (√2-2k)(√((1/2) + k√2 - k²) x - √(2k√2 - 2k² + x²) )

Fixed

(1/(k(k-√2))) ( (√2-2k)x(√((1/2) + k√2 - k²)) - √(2k√2 - 2k² + x²) )

I can go blind with these brackets!

Dr D, run that one by me again, how did you get the reciprocal x to show up in your calculations?

Okay, I see it now, you're using k as the approximating variable in the series expansion.

Answer 1

OK the brackets don't close but let's do it step by step.

As k --> 0,
(√2 - 2k)*√[1/2 + k√2 - k^2]*x
= x ...(1)

Now expanding binomially
(x^2 + 2√2 k - 2k^2)^(1/2)
= x + (√2 k - k^2)/x ...(2)

(1) - (2) = -(√2 k - k^2)/x
= (k^2 - √2 k) / x = k*(k - √2) / x

We need to multiply this result by the
1/[k*(k - √2)] at the beginning

We get 1/x

*EDIT*
Yes, the first part (1) is fairly straightforward as k--> 0.
In the second part (2), I"m using
(x^2 + a)^(1/2) = x + a/2x + ...
a = 2√2 k - 2k^2