# Thread: A limit that finally stumped me!

1. ## A limit that finally stumped me!

So anyone have any idea on how to solve this one? I've tried all the usual ways to solve..getting no where fast.

2. Hello,

The conjugate of a number a-b is a+b (and conversely). (a-b)(a+b)=aČ-bČ. This is why it can be very useful when you want to get rid of some square roots, or transforming a difference into a sum.

Multiply by $\displaystyle 1=\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}$

3. Ok - so I multiplied by the conjugate, as normal and I got:

$\displaystyle \frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}$

So I still zero out the denom as I come to 2...help here?

4. Originally Posted by 66replica
Ok - so I multiplied by the conjugate, as normal and I got:

$\displaystyle \frac{(\sqrt{6-x}-2)(\sqrt{3-x}-1)}{2-x}$

So I still zero out the denom as I come to 2...help here?
No, pardon me ! I misread and actually it was $\displaystyle \sqrt{3-x}+1$ for the conjugate !

Now, expand $\displaystyle (\sqrt{6-x}-2)(\sqrt{3-x}+1)$. In general, it must be a factor of x-2

5. Originally Posted by Moo
Hello,

The conjugate of a number a-b is a+b (and conversely). (a-b)(a+b)=aČ-bČ. This is why it can be very useful when you want to get rid of some square roots, or transforming a difference into a sum.

Multiply by $\displaystyle 1=\frac{\sqrt{3-x}-1}{\sqrt{3-x}-1}$

Small mistake:
Multiply by $\displaystyle 1=\frac{\sqrt{3-x}{\color{red}+}1}{\sqrt{3-x}{\color{red}+}1}$

To the OP: I'm very surprised that what moo showed you wasn't on your check list of all the usual ways ..... Clearly moo made a typo that you should have realised and corrected for.

6. Yeah it was a typo on my part. I've done the conjugate thing a few times on this problem and keep getting stuck trying to get an x-2 factored from the top.

I get stuck even after I foil the numerator I end up with:

$\displaystyle \frac{(\sqrt{6-x})(\sqrt{3-x})-2(\sqrt{3-x}+\sqrt{6-x}-2}{2-x}$

7. Originally Posted by 66replica
Yeah it was a typo on my part. I've done the conjugate thing a few times on this problem and keep getting stuck trying to get an x-2 factored from the top.

I get stuck even after I foil the numerator I end up with:

$\displaystyle \frac{(\sqrt{6-x})(\sqrt{3-x})-2(\sqrt{3-x}+\sqrt{6-x}-2}{2-x}$
You get $\displaystyle \frac{(\sqrt{6 - x} - 2)(\sqrt{3 - x} + 1)}{2-x}$.

Now multiply this expression by $\displaystyle \frac{(\sqrt{6 - x} + 2)}{\sqrt{6 - x} + 2}$.

You get $\displaystyle \frac{(2 - x) (\sqrt{3 - x} + 1)}{(2-x) (\sqrt{6 - x} + 2)} \, ....$

8. Thanks for the help - I understand the problem well now.

9. Wouldn't it be easier just to use L'hopital's rule? Then you would get $\displaystyle \frac{\sqrt{3-x}}{\sqrt{6-x}} | x=2 \Rightarrow 1/2$

10. Originally Posted by Hoover glasses
Wouldn't it be easier just to use L'hopital's rule? Then you would get $\displaystyle \frac{\sqrt{3-x}}{\sqrt{6-x}} | x=2 \Rightarrow 1/2$
Yes but most people don't learn L'hopital's rule until later in their calc study