# Evaluate Lim → 0, radicals in the numerator and denominator!

1. Feb 18, 2013

### physphys

1. The problem statement, all variables and given/known data

Evaluate lim x →0 √(x+1) - √(2x+1)
-----------------
√(3x+4) - √(2x+4)

2. Relevant equations

3. The attempt at a solution

First I rationalized the numerator by multiplying everything by √(x+1) + √(2x+1) / √(x+1) + √(2x+1)

so now i have
-x
----
( √(3x+4) - √(2x+4) )(√(x+1) + √(2x+1))

and I'm stuck from here, the answer is -2 so I'm not sure if I'm even right so far.
Thanks in advance for the help!

2. Feb 18, 2013

### jbunniii

Try multiplying the numerator and denominator by $\sqrt{3x+4}+\sqrt{2x+4}$?

3. Feb 18, 2013

### Staff: Mentor

Try L'Hôpital's rule.

4. Feb 18, 2013

### physphys

Well we never learned l'hopital's rule yet but after googling it and attempting it i got
(1/2) - 1
----------
(3/2) - (1/2)
which gave me -1/4, not the answer.

Now as for jbunnii's plan. I got up to
- x ( (√3x+1) + (√2x+4) )
------------------------
(x-3) ((√x+1) + (√2x+1))

I'm unsure of what to do next.

Edit: Can the X's cancel out here? so I end up with
- ( (√3x+1) + (√2x+4) )
------------------------
-3 ((√x+1) + (√2x+1))

Edit 2 : No that's giving me - 1/2, the inverse of what I need

5. Feb 18, 2013

### Staff: Mentor

How can you get -1/4? Anyway, the (3/2) in the denominator is wrong.

6. Feb 18, 2013

### jbunniii

How did you end up with $x-3$ in the denominator? I calculate
$$(\sqrt{3x+4} - \sqrt{2x+4})(\sqrt{3x+4} + \sqrt{2x+4}) = (3x+4) - (2x+4) = x$$
Now cancel the $x$ in the numerator and denominator and see what is left.

7. Feb 18, 2013

### jbunniii

Also, you somehow changed your $\sqrt{3x+4}$ into $\sqrt{3x+1}$.

8. Feb 18, 2013

### physphys

yeah you're right, I got the answer! I just copied the question down wrong. Thanks for the help everyone!

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