Rationalize this math problem...?

Question

[x+3] -[3]
--------------
x

the brackets represent radical signs

i need to rationalize this and find the limit...

the answer of the limit is 1/2[3] if that helps

Answer 1

To rationalize a binomial, multiply by its conjugate (which is the same two terms with a different sign between them). You'll come up with the square of one term minus the square of the other.

Unfortunately, you'll have to multiply the bottom by the same thing, which means that rationalizing the top will really get you nowhere.

Answer 2

Multiply top and bottom by √(x+3) + √3

you get √(x+3)² - √3² = x on top which cancels with the x on the bottom.

This leaves you with 1/(√(x+3) + √3) whose limit is 1/2√3

Answer 3

(?6 + ?3) / (?3 - ?6) ok so we are rationalizing, this is we are multiplying (?6 + ?3) / (?3 - ?6) with the help of (?3 + ?6) / (?3 + ?6), this could then be: ((?6 + ?3)(?3 + ?6)) / ((?3 - ?6)(?3 + ?6)) which upon enlargement and appearing multiplication and addition will provide: (6 + 9) / (3 - 9) which leads to: 15 / -6

Answer 4

([x+ 3]- [3])/x =
__ = ([x+ 3]- [3])([x+ 3] + [3])/{x ([x+ 3]+ [3]) }
__ = (x+ 3- 3)/{ x ([x+ 3]+ [3]) }
__ = x/{ x ([x+ 3]+ [3]) }
__ = 1/([x+ 3]+ [3])

Now you can apply the limit (looking at your answer, I suppose x tends to 0, doesn't it?)

__ lim 1/([x+ 3]+ [3]) when x tends to 0 = 1/(2[3]).

Answer 5

[sqr(x+3) -sqr(3) ] [sqr(x+3) +sqr(3)]
-------------------- . ------------------------
x [sqr(x+3) +sqr(3)]



x+3 -3
= --------------------------
x [sqr(x+3) +sqr(3)]


= x/{x [sqr(x+3) +sqr(3)]} =1/ sqr(x+3) +sqr(3)]

so the limit when x---->? which limit?