Evaluating Limit questions

[tpittman]

I have a question which involves a radical in the numerator. Normally this wouldn't be difficult, however I am confused about how to start this particular question:

lim (x --> 1) (√(2x^2-1) + x - 2) / (x^2 -1)

The issue is the -2 in the numerator, which leaves gives 3 terms in the numerator... How would I approach this problem?

Similarly, I am also having trouble beginning another question which contains a radical in both the numerator AND the denominator. I have attempted the conjugate method to rationalize the denominator, but simply ended up with a jumbled mess that I could not comprehend.

lim (x --> 4) (3 - √ (x+5)) / (√ (4x-7) - 3)

I appreciate any help I can get!
**I think it is also important to state that I am not allowed to use L'Hospital's rule, as my professor has not yet taught me this rule.

 

[srmichael]

I have a question which involves a radical in the numerator. Normally this wouldn't be difficult, however I am confused about how to start this particular question:

lim (x --> 1) (√(2x^2-1) + x - 2) / (x^2 -1)

The issue is the -2 in the numerator, which leaves gives 3 terms in the numerator... How would I approach this problem?

Similarly, I am also having trouble beginning another question which contains a radical in both the numerator AND the denominator. I have attempted the conjugate method to rationalize the denominator, but simply ended up with a jumbled mess that I could not comprehend.

lim (x --> 4) (3 - √ (x+5)) / (√ (4x-7) - 3)

I appreciate any help I can get!
**I think it is also important to state that I am not allowed to use L'Hospital's rule, as my professor has not yet taught me this rule.

For the first one, what are you using as your conjugate? If it's \(\displaystyle \sqrt(2x^2-1)+x+2\), that would be incorrect.

 

[tpittman]

I was attempting to use √(2x^2-1) - x - 2 as my conjugate, but honestly I am unsure of how to get the conjugate for this expression due to there being 3 terms instead of just 2.

 

[srmichael]

I was attempting to use √(2x^2-1) - x - 2 as my conjugate, but honestly I am unsure of how to get the conjugate for this expression due to there being 3 terms instead of just 2.

Treat (x-2) as one term and therefore use \(\displaystyle \sqrt(2x^2-1)-(x-2)\) as your conjugate. See what you get when you do this.

 

[tpittman]

This is still rather confusing to me. After multiplying by the conjugate, I have resulted in the following, which I am unsure as to it's validity...

2x^2-1 - (x-2)(x-2) / (x^2-1)((√2x^2-1)-x+2)

I assume that I can't simply cancel the (x^2-1) factor.



Okay, so after fully simplifying the numerator and then factoring the result, I managed to cancel the (x-1) term and ultimately solved the problem! Thank you for your assistance on the first problem..

 

[srmichael]

This is still rather confusing to me. After multiplying by the conjugate, I have resulted in the following, which I am unsure as to it's validity...

2x^2-1 - (x-2)(x-2) / (x^2-1)((√2x^2-1)-x+2)

I assume that I can't simply cancel the (x^2-1) factor.

OK, I see some effort on your part, so I'll help you out

I can't get LaTex to cooperate so here goes:

After multiplying by the conjugate the numerator will be (2x²-1) - (x-2)². When you expand (x-2)² and then combine like terms with (2x²-1), you get x² + 4x - 5 which factors to (x-1)(x+5). You will notice that in the denominator, x²-1 factors to (x-1)(x+1). You can then cancel out the (x-1) term and then proceed as normal with this limit.

 

[tpittman]

Thank you, I figured it out on my own moments before you posted!

Referring to the second question, I am now having difficulty with this question as well. I'm using the conjugate √4x-7 + 3 to rationalize the denominator, which I am confident is correct... Unfortunately, my basic math skills are failing me and I end up with a messy expression that I can't seem to simplify.

(3√4x-7 + 9 - [√x+5 * √4x-7] + 3√x+5) / 4x-16

Please let me know if this is incorrect, or how I may simplify it...

 

[pka]

2x^2-1 - (x-2)(x-2) / (x^2-1)((√2x^2-1)-x+2)
I assume that I can't simply cancel the (x^2-1) factor.
Okay, so after fully simplifying the numerator and then factoring the result, I managed to cancel the (x-1) term and ultimately solved the problem! Thank you for your assistance on the first problem.

The conjugate is \(\displaystyle \sqrt{x^2-1}+(x-2)\).
When you multiply the numerator becomes x^2+4x-5=(x+5)(x-1),

 

[tpittman]

The conjugate is \(\displaystyle \sqrt{x^2-1}+(x-2)\).
When you multiply the numerator becomes x^2+4x-5=(x+5)(x-1),

Thank you, I've managed to solve this problem. However, I am still having issues with the second part of my question, which can be viewed in my previous post!

 

[srmichael]

Thank you, I figured it out on my own moments before you posted!

Referring to the second question, I am now having difficulty with this question as well. I'm using the conjugate √4x-7 + 3 to rationalize the denominator, which I am confident is correct... Unfortunately, my basic math skills are failing me and I end up with a messy expression that I can't seem to simplify.

(3√4x-7 + 9 - [√x+5 * √4x-7] + 3√x+5) / 4x-16

Please let me know if this is incorrect, or how I may simplify it...

Yes, that is correct and yes, it is messy. You mentioned that you can't use L'Hospital's rule. Honestly, I'm not sure how you would do it otherwise. It is early for me so I may be missing something obvious, but even Wolfram uses L'Hospital's rule. Maybe one of the Usual Suspects (i.e. SK, Soroban, JeffM, Denis, etc.) may be able to provide their 2¢ worth and offer another way to do it without using L'Hospital's rule.

FWIW, the Wolfram shows the answer as -1/4.

 

[JeffM]

Thank you, I've managed to solve this problem. However, I am still having issues with the second part of my question, which can be viewed in my previous post!

I have tried a number of manipulations and not found one that gives a computable limit although numeric exploration suggests that the limit exists and is - 1/4. I feel stupid.

 

[lookagain]

Thank you, > > > I figured it out < < < on my own moments before you posted!

Referring to the second question, ...I am

Well, what value did you figure out (claim) for the first question?

 

[lookagain]

Referring to the second question, I am now having difficulty with this question as well.
I'm using the conjugate √(4x-7) + 3 to rationalize the denominator,
which I am confident is correct... Unfortunately, my basic math skills are failing me and I end up with a messy expression that I can't seem to simplify.

(3√(4x-7) + 9 - [√(x+5) * √(4x-7)] + 3√(x+5))/(4x-16)

Please let me know if this is incorrect, or how I may simplify it...

Yes, that is correct and yes, it is messy.

No, it is not correct when you don't have the needed grouping symbols.

 

[lookagain]

Maybe one of the Usual Suspects (i.e. SK, Soroban, JeffM, Denis, etc.)
may be able to provide their 2¢ worth and offer another way to do it without using L'Hospital's rule.

FWIW, the Wolfram shows the answer as -1/4.

I belong to the "etc." group.


\(\displaystyle \displaystyle\lim_{x\to \ 4} \ \dfrac{3 - \sqrt{x + 5}}{\sqrt{4x - 7} - 3}\)


Let x = y + 4


Then y = x - 4


\(\displaystyle As \ x \to 4, \ \ y \to 0.\)



\(\displaystyle \displaystyle\lim_{y\to \ 0} \ \dfrac{3 - \sqrt{y + 9}}{\sqrt{4y + 9} - 3}\)


or



\(\displaystyle \displaystyle\lim_{y\to \ 0} \ \dfrac{3 - \sqrt{9 + y}}{\sqrt{9 + 4y} - 3}\)


Using the Binomial Theorem, a square root method for polynomials, or other,
approximate it as follows:



The numerator can be \(\displaystyle \ 3 \ - \ \bigg(3 + \frac{1}{6}y - \frac{1}{216}y^2 \ + \ (the \ rest \ of \ terms \ have \ degree \ of \ at \ least \ 3)\bigg)\)



and the denominator is \(\displaystyle \ 3 + \frac{2}{3}y - \frac{2}{27}y^2 \ + \ (the \ rest \ of \ the \ terms \ have \ degree \ of \ at \ least \ 3) \ - \ 3\)



- - - - - - - - - - - - - - - - - - - - - - -



Then the numerator can be \(\displaystyle \ - \frac{1}{6}y + \frac{1}{216}y^2 \ + \ (the \ rest \ of \ the \ terms \ have \ degree \ of \ at \ least \ 3)\bigg)\)



and the denominator is \(\displaystyle \ \frac{2}{3}y - \frac{2}{27}y^2 \ + \ (the \ rest \ of \ the \ terms \ have \ degree \ of \ at \ least \ 3) \)



\(\displaystyle \displaystyle\lim_{y\to \ 0} \ \dfrac{\frac{-1}{6}y + \frac{1}{216}y^2 \ + \ 3rd \ degree \ or \ higher \ terms}{\frac{2}{3}y - \frac{2}{27}y^2 \ + \ 3rd \ degree \ or \ higher \ terms} \)



\(\displaystyle \displaystyle\lim_{y\to \ 0} \ \dfrac{y(\frac{-1}{6} + \frac{1}{216}y \ + \ 2nd \ degree \ or \ higher \ terms)}{y(\frac{2}{3} - \frac{2}{27}y \ + \ 2nd \ degree \ or \ higher \ terms)}\)




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