Evaluate Lim → 0, radicals in the numerator and denominator

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Homework Help Overview

The problem involves evaluating the limit as x approaches 0 of a fraction with square roots in both the numerator and denominator. The original poster attempts to simplify the expression by rationalizing the numerator.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss rationalizing the numerator and suggest multiplying by the conjugate. Some mention using L'Hôpital's rule, although one participant notes they have not learned it yet. There are questions about the accuracy of calculations and the simplification process.

Discussion Status

The discussion is active, with participants providing different approaches and checking each other's calculations. There is no explicit consensus, but some guidance has been offered regarding the rationalization and simplification steps.

Contextual Notes

One participant mentions not having learned L'Hôpital's rule, which may affect their approach to the problem. There are also indications of confusion regarding the calculations and the setup of the problem.

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Homework Statement



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

Homework Equations





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!
 
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Try multiplying the numerator and denominator by ##\sqrt{3x+4}+\sqrt{2x+4}##?
 
Try L'Hôpital's rule.
 
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
 
physphys said:
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.

How can you get -1/4? Anyway, the (3/2) in the denominator is wrong.
 
physphys said:
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))
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.
 
Also, you somehow changed your ##\sqrt{3x+4}## into ##\sqrt{3x+1}##.
 
jbunniii said:
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.

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

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