Limits of Infinity

[KidInkFan]

Hi, just need help starting the question:

The lim as x approaches negative infinty of (6x-7)/squareroot of (4x^2+11)

Do i rationalize the numerator? Thank you for any help!!

 

[lookagain]

Hi, just need help starting the question:

The lim as x approaches negative infinty of (6x-7)/squareroot of (4x^2+11)

Do i rationalize the numerator? Thank you for any help!!

The numerator is already rationalized. Also, I don't see
any advantage (if it were to be brought up) with
multiplying the numerator and denominator by the
conjugate of the numerator.


Edit:

One of the ways that I can consider is to let

x = -y. As x approaches -oo, so does -y.

Then y approaches +oo.


The problem can become lim (y --> oo) of (-6y - 7)/(sqrt(4x^2 + 11)).

 

[JeffM]

Hi, just need help starting the question:

The lim as x approaches negative infinty of (6x-7)/squareroot of (4x^2+11)

Do i rationalize the numerator? Thank you for any help!!

I do not see that rationalizing the denominator will help on THIS problem, but it is not a silly line of thought. On problems of this kind, you may have to experiment with a few ideas before finding one that will let you transform the expression into an equivalent one that is tractable. Let's get this so both numerator and denominator are in square root form.

\(\displaystyle Make\ explicit\ the\ negativity\ of\ x\ in\ the\ limit \implies \dfrac{6x - 7}{\sqrt{4x^2 + 11}} = - \dfrac{-(6x + 7)}{\sqrt{4x^2 + 11}}.\)

\(\displaystyle Put\ numerator\ and\ denominator\ in\ square\ root\ form \implies - \dfrac{\sqrt{\left\{-(6x + 7)\right\}^2}}{\sqrt{4x^2 + 11}} = - \dfrac{\sqrt{36x^2 + 84x + 49}}{\sqrt{4x^2 + 11}} = - \sqrt{\dfrac{36x^2 + 84x + 49}{4x^2 + 11}}.\)

What next?

 

[lookagain]

Like lookagain, I do not see that rationalizing the denominator will help on THIS problem,

> > but it is not a silly line of thought. < <

On problems of this kind, you may have to experiment with a few ideas before finding one that will let you
transform the expression into an equivalent one that is tractable.

As mmm4444bot, would point out, do not put words in my mouth. I did not state that it is a "silly line of thought." <--- edit
You are stating a disagreement to something I did not state.

 

[JeffM]

As mmm4444bot, would point out, do not put words in my mouth. I did not state that it is "silly."
You are stating a disagreement to something I did not state.

I did not put any words into your mouth. Please show me where I said that you said anything was silly. When someone goes down a wrong path, it is sometimes reassuring to them to point out that it was not obviously wrong or that taking a wrong path is nothing to worry about. The student's problem here is that when one path proved fruitless, he had no idea where to go and stopped. Encouraging people to try different paths and to avoid discouragement when one path does not work are not things that I am going to stop doing.

I shall edit my original post, however, to delete my acknowledgement that you were the first to see that rationalizing the denominator was not going to be helpful.

 

[lookagain]

I did not put any words into your mouth. Please show me where I said that you said anything was silly.
When someone goes down a wrong path, it is sometimes reassuring to them to point out that it was not obviously wrong or
that taking a wrong path is nothing to worry about. The student's problem here is that when one path proved fruitless,
he had no idea where to go and stopped. Encouraging people to try different paths and to avoid discouragement when one
path does not work are not things that I am going to stop doing.

I shall edit my original post, however, to delete my acknowledgement that you were the first to see that rationalizing the
denominator was not going to be helpful.

The student asked about rationalizing the numerator, not the denominator. I never addressed about
rationalizing the denominator, so why did you bring that up?

Here is the pertinent portion of which I stated:

"Also, I don't see any advantage (if it were to be brought up) with
multiplying the numerator and denominator by the
conjugate of the numerator."


JeffM stated: "I shall edit my original post, however, to delete my acknowledgement that you were the
first to see that rationalizing the denominator was not going to be helpful."

1) I didn't refer to the denominator. I addressed the student referring to the numerator. I did not
mention anything about rationalizing the denominator. The mention of the denominator
was about (the appropriateness)of multiplying it and the numerator by the conjugate of the numerator.


2) I didn't state that (whatever) "wasn't going to be helpful." I stated that "I don't see
any advantage." I don't claim to know as a surety if it won't be helpful or not.

 

[JeffM]

The student asked about rationalizing the numerator, not the denominator. I never addressed about
rationalizing the denominator, so why did you bring that up?

Here is the pertinent portion of which I stated:

"Also, I don't see any advantage (if it were to be brought up) with
multiplying the numerator and denominator by the
conjugate of the numerator."


JeffM stated: "I shall edit my original post, however, to delete my acknowledgement that you were the
first to see that rationalizing the denominator was not going to be helpful."

1) I didn't refer to the denominator. I addressed the student referring to the numerator. I did not
mention anything about rationalizing the denominator. The mention of the denominator
was about (the appropriateness)of multiplying it and the numerator by the conjugate of the numerator.


2) I didn't state that (whatever) "wasn't going to be helpful." I stated that "I don't see
any advantage." I don't claim to know as a surety if it won't be helpful or not.

I shall leave it up to the student to determine what reply was most helpful to him or her.

Have a nice day.

 

[HallsofIvy]

As mmm4444bot, would point out, do not put words in my mouth. I did not state that it is a "silly line of thought." <--- edit
You are stating a disagreement to something I did not state.

I can see no evidence that JeffM was responding to you post. I see that you have both edited your posts. Was something relevant posted?

 

[JeffM]

I can see no evidence that JeffM was responding to you post. I see that you have both edited your posts. Was something relevant posted?

I deleted my reference to lookagain's original post as I said I would. You can see what the original post was because lookagain quotedthe relevant portion.

 

[lookagain]

\(\displaystyle - \dfrac{-(6x + 7)}{\sqrt{4x^2 + 11}}.\)

\(\displaystyle Put\ numerator\ and\ denominator\ in\ square\ root\ form \implies - \dfrac{\sqrt{\left\{-(6x + 7)\right\}^2}}{\sqrt{4x^2 + 11}} \)

\(\displaystyle - \dfrac{-(6x + 7)}{\sqrt{4x^2 + 11}} \ \ne \ - \dfrac{\sqrt{\left\{-(6x + 7)\right\}^2}}{\sqrt{4x^2 + 11}} \)


\(\displaystyle - \dfrac{\sqrt{\left\{-(6x + 7)\right\}^2}}{\sqrt{4x^2 + 11}} \ = \ - \dfrac{|-(6x + 7)|}{\sqrt{4x^2 + 11}}, \ \ just \ \ as \ \ \sqrt{x^2} \ = \ |x|. \)

 

[JeffM]

\(\displaystyle - \dfrac{-(6x + 7)}{\sqrt{4x^2 + 11}} \ \ne \ - \dfrac{\sqrt{\left\{-(6x + 7)\right\}^2}}{\sqrt{4x^2 + 11}} \)


\(\displaystyle - \dfrac{\sqrt{\left\{-(6x + 7)\right\}^2}}{\sqrt{4x^2 + 11}} \ = \ - \dfrac{|-(6x + 7)|}{\sqrt{4x^2 + 11}}, \ \ just \ \ as \ \ \sqrt{x^2} \ = \ |x|. \)

Given that the problem is about the limit as x approaches minus infinity and that the previous step had mentioned that aspect of the problem, the assumption that x < - (7/6) is warranted.

\(\displaystyle x < - \dfrac{7}{6} \implies 6x < - 7 < 6x + 7 < 0 \implies 0 < -(6x + 7) = \sqrt{36x^2 + 84x + 49} = \sqrt{\left\{-(6x + 7)\right\}^2}.\)

Perhaps I am misunderstanding your point. I shall of course correct my post if there is an error.

 

[KidInkFan]

Ahh I think i understand and once i get the the whole equation as a square root function i can take out the common denominator which is x^2 leaving me with an answer of negative square root of 36/4?

 

[JeffM]

Ahh I think i understand and once i get the the whole equation as a square root function i can take out the common denominator which is x^2 leaving me with an answer of negative square root of 36/4?

This is not expressed correctly although you may be thinking along the right track.

You now have the square root of a rational fraction with the numerator and denominator of the same degree.

\(\displaystyle - \sqrt{\dfrac{36x^2 + 84x + 49}{4x^2 + 11}} = - \sqrt{9 + \dfrac{84x - 50}{4x^2 + 11}}.\) Nothing to do with common denominators.

Now when I take the limit of that I get

\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}- \sqrt{9 + \dfrac{84x - 50}{4x^2 + 11}} = - \lim_{x \rightarrow \infty}\sqrt{9 + \dfrac{84x}{4x^2}} = - \lim_{x \rightarrow \infty}\sqrt{9 + \dfrac{21}{x}}= - \lim_{x \rightarrow \infty}\sqrt{9 + 0} = - 3.\)

 

[lookagain]

I do not see that > > rationalizing the denominator ,< < will help on THIS problem, ...

Suppose I decided to use that method anyway:


\(\displaystyle \displaystyle\lim_{x \to -\infty} \ \dfrac{6x - 7}{\sqrt{4x^2 + 11}} \ = \)


Let x = -y.

As x --> -oo, -y --> -oo.

Then, as x --> -oo, y --> +oo.


\(\displaystyle \displaystyle\lim_{y \to +\infty} \ \dfrac{6(-y) - 7}{\sqrt{4(-y)^2 + 11}} \ = \)


\(\displaystyle \displaystyle\lim_{y \to +\infty} \ - \dfrac{6y + 7}{\sqrt{4y^2 + 11}} \)



If the denominator were to be rationalized at this point, the fraction would become:



\(\displaystyle \displaystyle\lim_{y \to +\infty} \ - \dfrac{(6y + 7) \sqrt{4y^2 + 11}}{4y^2 + 11} \ = \)


Multiply the numerator and the denominator by \(\displaystyle \dfrac{1}{y^2}, \) but split it up in the numerator:


\(\displaystyle \displaystyle\lim_{y \to +\infty} \ - \dfrac{\bigg(\dfrac{1}{y}\bigg)(6y + 7)\bigg (\dfrac{1}{y}\bigg)\sqrt{4y^2 + 11}}{\bigg(\dfrac{1}{y^2}\bigg)(4y^2 + 11)} \ = \)



Square \(\displaystyle \dfrac{1}{y}, \) and bring it inside the radicand as a factor:



\(\displaystyle \displaystyle\lim_{y \to +\infty} \ - \dfrac{\bigg(\dfrac{1}{y}\bigg)(6y + 7) \sqrt{\dfrac{1}{y^2}\bigg(4y^2 + 11\bigg)}}{\bigg(\dfrac{1}{y^2}\bigg)(4y^2 + 11)} \ = \)





\(\displaystyle \displaystyle\lim_{y \to +\infty} \ - \dfrac{\bigg(6 + \dfrac{7}{y}\bigg) \sqrt{4 + \dfrac{11}{y^2}}}{4 + \dfrac{11}{y^2}} \ = \)




\(\displaystyle - \dfrac{(6 + 0) \sqrt{4 + 0}}{4 + 0} \ = \)



\(\displaystyle - \dfrac{(6)(2)}{4} \ = \)



\(\displaystyle \boxed{ \ -3 \ }\)

 

[KidInkFan]

Oh right sorry I didnt mean common denominator, I meant the same degree but thank you for explaining the question in much more detail!! i really apreciate the help!!