Evaluate Lim → 0, radicals in the numerator and denominator

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The discussion focuses on evaluating the limit as x approaches 0 for the expression involving square roots in both the numerator and denominator. Participants suggest rationalizing the numerator and using techniques like L'Hôpital's rule, although one user notes they haven't learned it yet. Confusion arises regarding the calculations, particularly with the denominator, where errors in transcription lead to incorrect conclusions. Ultimately, after clarification and corrections, one user successfully finds the correct answer by realizing they had copied the original problem incorrectly. The thread highlights the importance of careful notation and step-by-step verification in solving limits involving radicals.
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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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