find the limit of (sqrt(2-x)-sqrt(2))/x as x->0?

Question

Without using L'Hostpital's rule...

This is asked as a simple question in introductory calculus. I have done loads of calculus and am starting to feel stupid that I cannot get this. I have tried (to the extent of wasting trees) and googled/wolfram'd the answer. I know the answer is -1/2(sqrt2) but its the why that is annoying me! Wolfram uses L'Hostpital's rule... I'm sure it's reeeeally simple?

Help?

Update:

ah! Conjugate, then apply limit laws as per usual... right... got it... obviously I'm going mad =D

Answer 1

Favorite Answer

(lim x → 0) (√(2 - x) - √2)/x

= (√(2 - x) - √2)(√(2 - x) + √2) / (x (√(2 - x) + √2))

= ((2 - x) - 2) / (x (√(2 - x) + √2))

= -x / (x (√(2 - x) + √2))

= -1 / (√(2 - x) + √2)) Applying limit

= -1 / (√(2 - 0) + √2)

= -1 / (2√2)

= (-1/2)√2.

No, my answer is right: http://www.wolframalpha.com/input/?i=limit+%28%28s...

Answer 2

the conjugate quantity is sqrt(2-x) + sqrt(2)

then multiply and divide by the conjugate quantity.

By the way the answer is - (1/4) * sqrt(2)

You should put the parentheses in your answer

-1/2(sqrt2) should be -1/ [ 2 sqrt(2) ]

which can be simplified to my answer - (1/4) * sqrt(2)

or - sqrt(2) /4 to avoid confusion.