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

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?

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

### 2 Answers

- Anonymous1 decade agoFavorite 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...

- nleLv 71 decade ago
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.