Calculus question about rationalizing a fraction

So the problem is:

If f is continuous at x=2, and if f(x)=√(x+2) - √(2x)/(x-2) for x≠ 2
k for x=2
then what is the value of k?

Sorry if that's hard to understand....in words, if f is continuous at x=2, and if f(x) is equal to the square root of x+2 - the square root of 2x all over x-2 for x≠2, and if f(x) is equal to k for x=2, then k = ?

Anyway, I know how to set up the problem, calc-wise. I just need help rationalizing this fraction. I multiplied the whole thing by the conjugate of the numerator and winded up with x + 2 - 2x/(x-2)(√x+2 + √2x). I can't figure out how to simplify it from there.

For the total answer to the problem I got k=1/4, but apparently k is supposed to equal -1/4...so any help figuring out where I went wrong would be appreciated!
Thanks!

f(x)=[sqrt(x+2)-sqrt(2x)]/(x-2)
For f to be continuous at x=2, we must have
f(2)=lim{x-->2} f(x)
To determine this limit, I multiply and divide by the quantity (sqrt(x+2)+sqrt(2x))
[(sqrt(x+2)-sqrt(2x))(sqrt(x+2)+sqrt(2…
In the numerator, the product is a difference of two squares
[(x+2)-(2x)]/[(x-2)*(sqrt(x+2)+sqrt(2x…
[-x+2]/[(x-2)*(sqrt(x+2)+sqrt(2x))]
If I factor out a -1 in the numerator, we get
[(-1)(x-2)]/[(x-2)(sqrt(x+2)+2x)]
Canceling the common (x-2) factors
-1/[sqrt(x+2)+sqrt(2x)]
Now take limit as x-->2
limit{x-->2} (-1/[sqrt(x+2)+sqrt(2x)]= -1/[sqrt(2+2)+sqrt(2*2)]
= -1/(2+2)= -1/4
Hope this Helps!

Do you know l'Hopitale's rule? The limit of a(x)/b(x) at x = 2 is the same as the limit of da(x)/dx / db(x)/x

Conveniently d(x-2)/dx= 1, so the answer is d(sqrt(x+2)-sqrt(2x))/dx at x = 2

Do the differentiation and you'll get the answer, I agree its -1/4