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\[\lim_{x \rightarrow 2}\frac{ \sqrt{6-x}-2 }{\sqrt{3-x}-1 }\]

You can multiply the bottom by the conjugate or since the value is 0/0, you can use L'Hopital's rule

\[\frac{\sqrt{18-9x+x^2}+\sqrt{6-x}-2\sqrt{3-x}-2}{(3-x) - 1}\]

Is that what it should work out to? :/

I'll show you what I did the first time, and got points off for:

That algebra is in fact wrong, correct, at least the 1st term of the numerator in the product?

Damn, alright...

I'm going to take a minute to work through it if that's okay?

Is this a rule for any situations with a radical in both the numerator and denominator?

not that I can remember, but this trick of using 2 conjugates seems to work

\[\Large (a+b)*(c+d) \ne a*c + b*d\]

you have to use the FOIL rule

ok go ahead

Can I cancel out that string of common terms in the top and bottom?

Sorry, in mine?

I'm not sure what you mean. Have a look at my steps. Let me know if they make sense or not

That just makes the limit, or answer, go to zero then... that can't be right.

Yep, that's what I did:)

I don't get what to do then!