Lim x→0 ((√2 + x)−(√2 − x)) / x

I am assuming you want to find:
lim (x-->0) [√(2 + x) - √(2 - x)]/x.

If so, we need to rationalize the numerator; to do this, we need to multiply the numerator and denominator by the conjugate of the numerator, which is √(2 + x) + √(2 - x). Doing so yields:
lim (x-->0) [√(2 + x) - √(2 - x)]/x
= lim (x-->0) [√(2 + x) + √(2 - x)][√(2 + x) - √(2 - x)]/{x[√(2 + x) + √(2 - x)]}
= lim (x-->0) [(2 + x) - (2 - x)]/{x[√(2 + x) + √(2 - x)]}, by applying difference of squares
= lim (x-->0) 2x/{x[√(2 + x) + √(2 - x)]}, by simplifying the numerator
= lim (x-->0) 2/[√(2 + x) + √(2 - x)], by canceling x
= 2/[√(2 + 0) + √(2 - 0)], by evaluating the result at x = 0
= 2/(2√2)
= √2/2.

I hope this helps!

1/√2 and √2/2 are the same thing, but √2/2 is rationalized.

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