Evaluate this limit with square roots

[aeromat]

Homework Statement

$<br /> \lim_{x\to0}\frac{\sqrt{x+1} - \sqrt{2x+1}}{\sqrt{3x + 4} - \sqrt{2x+4}}<br />

The Attempt at a Solution

Ok, I just want to know what is the easiest approach tactic one could take to solve this. I tried doing the conjugate to rationalize the denominator, but then I am left with a gigantic numerator on to the top with many square root multiples.

 

[Mark44]

Homework Statement

$<br /> \lim_{x\to0}\frac{\sqrt{x+1} - \sqrt{2x+1}}{\sqrt{3x + 4} - \sqrt{2x+4}}<br />

The Attempt at a Solution

Ok, I just want to know what is the easiest approach tactic one could take to solve this. I tried doing the conjugate to rationalize the denominator, but then I am left with a gigantic numerator on to the top with many square root multiples.

You could use L'Hopital's Rule on this one. Have you tried that?

 

[flyingpig]

rationalize to a sum of squares

 

[Mark44]

rationalize to a sum of squares

I think you mean "difference of squares."

 

[flyingpig]

Yes that one lol

 

[aeromat]

Ok what if I haven't learned L'Hopital's rule, what strategy could I take?

 

[aeromat]

Ok nevermind, I am trying to use the L'Hopital's Rule:

If I want to express f(x) = \sqrt{x+1} - \sqrt{2x+1} using the Power Rule, how would I simplify it with the "-1/2" exponent?

 

[Mark44]

It would be an exponent of +1/2, not -1/2.

f(x) = (x + 1)^1/2 - (2x - 1)^1/2

Is that what you're asking?

 

[Mark44]

L'Hopital's Rule worked for me - I was able to get a number for this limit. Multiplying numerator and denominator by the conjugate of the denominator did not work for me, as the resulting expression is still in the indeterminate form [0/0].

 

[aeromat]

Wait, why would it not be -1/2? When you apply the Power Rule, you do n-1 for the exponent, and in this case it was 1/2. So 1/2 - 1 should be -1/2, wouldn't it be?

 

[Dickfore]

Hint:

<br /> (1 + x)^{\alpha} \sim 1 + \alpha x, \; x \rightarrow 0<br />

With this formula, you can find the asymptotic behavior of the numerator and denominator. I'll do one for you:

<br /> \sqrt{3 x + 4} = \left[4 \left(1 + \frac{3 x}{4}\right)\right]^{\frac{1}{2}} \sim 2 \, \left(1 + \frac{1}{2} \, \frac{3 x}{4}\right) = 2 + \frac{3 x}{4}<br />

and so on. You will see the leading terms will cancel separately in the numerator and the denominator. That is why you need to go to next to leading order. Then, you will see a nice cancellation.

 

[Dick]

When you rationalize at the first stage you multiply numerator and denominator by sqrt(3x+4)+sqrt(2x+4), yes? That gives you x in the denominator and it's still 0/0. That's ok, but you don't need to leave the complicated sqrt(3x+4)+sqrt(2x+4) in the numerator. You know that limit is 4. Just multiply the numerator by 4. Now use the same strategy when you rationalize the numerator. Keep things whose limits are nonzero constants as constants until you need to expand them. It doesn't have to be a sqrt nightmare.

 

[Mark44]

It would be an exponent of +1/2, not -1/2.

f(x) = (x + 1)^1/2 - (2x - 1)^1/2

Is that what you're asking?

Wait, why would it not be -1/2? When you apply the Power Rule, you do n-1 for the exponent, and in this case it was 1/2. So 1/2 - 1 should be -1/2, wouldn't it be?

f(x) above is the numerator of your expression before applying L'Hopital's Rule. After applying this rule the numerator is f'(x) = (1/2)(x + 1)^-1/2 - (2/2)(2x - 1)^-1/2.

The question you asked in post #7 was unclear to me, which is why I asked for clarification on what you were trying to say.

 

[aeromat]

1/(2√[x+1]) - 1/(2√[2x+1])
---------------------------------
1/2(√[3x+4]) - 1/2(√[2x+4])

That is what I got after taking the derivative of both. Now I am stuck, because if I were to plug in x, I get a 0/0 indeterminate form again.

 

[Mark44]

You have not used the chain rule correctly.

For example,
\frac{d}{dx} (2x + 1)^{1/2} = \frac{2}{2(2x + 1)^{1/2}}= \frac{1}{(2x + 1)^{1/2}}

You have mistakes in three of your derivatives.

 

[aeromat]

You have not used the chain rule correctly.

For example,
\frac{d}{dx} (2x + 1)^{1/2} = \frac{2}{2(2x + 1)^{1/2}}= \frac{1}{(2x + 1)^{1/2}}

You have mistakes in three of your derivatives.

Thank you very much. I finally got the answer, -2.

 

[Mark44]

Thank you very much. I finally got the answer, -2.

Yep, that's what I got. Good job!

 

[Dickfore]

http://www.wolframalpha.com/input/?i=Limit[(Sqrt[x+1]-Sqrt[2*x+1])/(Sqrt[3*x+4]-Sqrt[2*x+4]),x->0]