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\[\frac{ 4+h-4 }{ h(\sqrt{4+h}+2) }=\frac{ 1 }{\sqrt{4+h}+2 }\]
nomarlly limit question

we multiplied by\[\sqrt{4+h}+2\]
both num and denominator

hmmm...isn't the point of rationalization to remove radicals from the denominator?

Well then there were no radicals in the denominator

So u wld keep it the way it is

really?

imagine if this question was\[\lim_{h \rightarrow 0}\frac{ \sqrt{4+h}-2 }{ h }\]

??

you cant just plug h=0 here but in the rationalised one

Ohh that is cool :P

limits in an algebra question? that's morbid.....

Rationalizing means to remove the radical from the place it is in...

hmmm well u can be asked to rationalize the numerator too

now im confused with the contradictions...

What are the contradictions?

@swissgirl said don't change...now she says change

Just use ur own brain -_-

lol

someone's wrong here....wonder who

\[\frac{ 1 }{ \sqrt{2} }=\frac{ \sqrt{2} }{ 2 }\]

rationalised

i know what happens if it's in the denominator

trust me lgba knows how to rationalize a numerator or denomanator

the question is if it's in the numerator

See how you can't remove the radical from both numerator and denominator?

Yes, so move it to the denominator. Do not care about the denominator.

my question is not how to rationalize @swissgirl but if it's suppose to be rationalized

You can rationalize it, but mathematicians always love if the radical is in the numerator.

because what i know is that if it's in the numerator, then it's okay

It's better to put radicals in the numerator rather than the denominator.

so why put in denominator then?

Yup, I saw that you could rationalize the numerator too though it's rare.

when i went online, all i saw were unreliable sources on rationalizing numerators

http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut41_rationalize.htm

Never rationalize the numerator unless given a question to perform.

http://tutorial.math.lamar.edu/Extras/AlgebraTrigReview/Rationalizing.aspx

These are both reliable sources

hmm

the links win. can't argue with that.

\[\frac{ \sqrt{x+h}-\sqrt{x} }{ h }\]ration.
\[\frac{ 1 }{ \sqrt{x+h}+\sqrt{x} }\]

square roots in denominators really look weird...