Limit, x -> infty, of sqrt(x^2 + ax) - sqrt(x^2 + bx)

[mmm4444bot]

Good Day!

I'm considering the following limit.

$\displaystyle \lim_{x \to \infty} \sqrt{x^2 + a \cdot x} \;-\; \sqrt{x^2 + b \cdot x}$

I used algebra to arrive at the following.

$\displaystyle \lim_{x \to \infty} \sqrt{x} \cdot \frac{a - b}{\sqrt{x + a} \;+\; \sqrt{x + b}}$

To proceed from this point, I reasoned that the parameters become insignificant as x becomes infinite.

$\displaystyle \lim_{x \to \infty} \sqrt{x} \cdot \frac{a - b}{\sqrt{x + 0} \;+\; \sqrt{x + 0}}$

$\displaystyle \lim_{x \to \infty} \sqrt{x} \cdot \frac{a - b}{2 \cdot \sqrt{x}} \;=\; \frac{a - b}{2}$

I believe that the result is correct, but I'm not particularily satisfied with having to include the analysis regarding the parameters.

Is anybody able to eliminate the variable algebraically or otherwise?

(Besides using the conjugate, I experimented with completing the square and also dividing the radicands by x^2 -- all to no avail. I'm stumped.)

Cheers,

~ Mark

 

[pka]

Re: Limit Exercise

You need to check your algebra.
$\displaystyle \sqrt {x^2 + ax} - \sqrt {x^2 + bx} = \frac{{ax - bx}}{{\sqrt {x^2 + ax} + \sqrt {x^2 + bx} }} = \frac{{a - b}}{{\sqrt {1 + \frac{a}{x}} + \sqrt {1 + \frac{b}{x}} }}$

 

[Deleted member 4993]

Re: Limit Exercise

Good Day!

I'm considering the following limit.

$\displaystyle \lim_{x \to \infty} \sqrt{x^2 + a \cdot x} \;-\; \sqrt{x^2 + b \cdot x}$the parameters become insignificant as x becomes infinite. -- If we apply that reason here - the limit becomes 0.


I used algebra to arrive at the following.

$\displaystyle \lim_{x \to \infty} \sqrt{x} \cdot \frac{a - b}{\sqrt{x + a} \;+\; \sqrt{x + b}}$

To proceed from this point, I reasoned that the parameters become insignificant as x becomes infinite.

$\displaystyle \lim_{x \to \infty} \sqrt{x} \cdot \frac{a - b}{\sqrt{x + 0} \;+\; \sqrt{x + 0}}$

$\displaystyle \lim_{x \to \infty} \sqrt{x} \cdot \frac{a - b}{2 \cdot \sqrt{x}} \;=\; \frac{a - b}{2}$

I believe that the result is correct, but I'm not particularily satisfied with having to include the analysis regarding the parameters.

Is anybody able to eliminate the variable algebraically or otherwise?

(Besides using the conjugate, I experimented with completing the square and also dividing the radicands by x^2 -- all to no avail. I'm stumped.)

Cheers,

~ Mark

 

[PAULK]

Re: Limit Exercise

You need to check your algebra.
$\displaystyle \sqrt {x^2 + ax} - \sqrt {x^2 + bx} = \frac{{ax - bx}}{{\sqrt {x^2 + ax} + \sqrt {x^2 + bx} }} = \frac{{a - b}}{{\sqrt {1 + \frac{a}{x}} + \sqrt {1 + \frac{b}{x}} }}$

No; the algebra is correct (nothing wrong with yours, either) but I agree -- yours is more transparent.

 

[mmm4444bot]

Re: Limit Exercise

You need to check your algebra.
$\displaystyle \sqrt {x^2 + ax} - \sqrt {x^2 + bx} = \frac{{ax - bx}}{{\sqrt {x^2 + ax} + \sqrt {x^2 + bx} }} = \frac{{a - b}}{{\sqrt {1 + \frac{a}{x}} + \sqrt {1 + \frac{b}{x}} }}$

Thank you, PKA.

I tried to get the variable into the denominator of fractions (just as you did) on one of my first attempts by multiplying the original expression by sqrt(x^2)/sqrt(x^2). But, of course, that led to sqrt(1) - sqrt(1).

I'll try to obtain your result.

Cheers,

~ Mark

 

[mmm4444bot]

Re: Limit Exercise

... the parameters become insignificant as x becomes infinite. -- If we apply that reason here - the limit becomes 0.

Hi Subhotosh:

Good grief. Right you are!

This type of sloppiness has bitten me before -- in that I do not always go back to the beginning to verify a result when I make assumptions. I hope I remember your lesson ...

Cheers,

~ Mark

 

[mmm4444bot]

Re: Limit Exercise

... No; the algebra is correct (nothing wrong with yours, either) but I agree -- yours is more transparent.

Hello Paulk:

Thank you for your confirmation. I worked this exercise from a few different approaches, so that by the time I arrived at my posted result I was confident that the algebra is sound.

I simply did not arrive at the result I wanted, which is the form that PKA posted.

Can you identify the flaw in the following?

$\displaystyle \sqrt{x^2 + a \cdot x} \;-\; \sqrt{x^2 + b \cdot x}$

$\displaystyle \frac{\sqrt{x^2}}{\sqrt{x^2}} \cdot \left(\sqrt{x^2 + a \cdot x} \;-\; \sqrt{x^2 + b \cdot x}\right)$

$\displaystyle \sqrt{x^2} \cdot \left(\frac{\sqrt{x^2 + a \cdot x}}{\sqrt{x^2}} \;-\; \frac{\sqrt{x^2 + b \cdot x}}{\sqrt{x^2}}\right)$

$\displaystyle \sqrt{x^2} \cdot \left(\sqrt{1 + \frac{a}{x}} \;-\; \sqrt{1 + \frac{b}{x}}\right)$

Taking the limit on this expression leads to 0.

Cheers,

~ Mark

 

[Deleted member 4993]

Re: Limit Exercise

... No; the algebra is correct (nothing wrong with yours, either) but I agree -- yours is more transparent.

Hello Paulk:

Thank you for your confirmation. I worked this exercise from a few different approaches, so that by the time I arrived at my posted result I was confident that the algebra is sound.

I simply did not arrive at the result I wanted, which is the form that PKA posted.

Can you identify the flaw in the following?

$\displaystyle \sqrt{x^2 + a \cdot x} \;-\; \sqrt{x^2 + b \cdot x}$

$\displaystyle \frac{\sqrt{x^2}}{\sqrt{x^2}} \cdot \left(\sqrt{x^2 + a \cdot x} \;-\; \sqrt{x^2 + b \cdot x}\right)$

$\displaystyle \sqrt{x^2} \cdot \left(\frac{\sqrt{x^2 + a \cdot x}}{\sqrt{x^2}} \;-\; \frac{\sqrt{x^2 + b \cdot x}}{\sqrt{x^2}}\right)$

$\displaystyle \sqrt{x^2} \cdot \left(\sqrt{1 + \frac{a}{x}} \;-\; \sqrt{1 + \frac{b}{x}}\right)$

Taking the limit on this expression leads to 0. - No it is indeterminate --> 0 times infinity Cheers,

~ Mark

 

[daon]

Re: Limit Exercise

I used algebra to arrive at the following.

$\displaystyle \lim_{x \to \infty} \sqrt{x} \cdot \frac{a - b}{\sqrt{x + a} \;+\; \sqrt{x + b}}$

I simply did not arrive at the result I wanted, which is the form that PKA posted.

Take out $\displaystyle \sqrt{x}$ from the denominator of the result you did arrive at.

 

[mmm4444bot]

Re: Limit Exercise

- No it is indeterminate --> 0 times infinity

Hi Subhotosh:

Would you say that the following statement is correct?

"When multiplying a 'number' with infinite magnitude by zero, the result is unknown."

In other words, we say that the Zero-Product Property of Real Numbers does not apply when infinity is involved.

I understand that numbers with infinite magnitude cannot be "nailed down" as any specific real number, but (until now) I was thinking that a factor of zero would result in a product of zero, nonetheless.

(The concept of infinity was not discussed in any detail during my undergraduate years; nor do I see any such discussion in current precalculus texts. Perhaps precalculus courses should include a section devoted specifically to infinity, as opposed to having students pick up bits and pieces of information as they move on through calculus.)

Anyway, thank you for the continued discussion.

Cheers,

~ Mark

 

[mmm4444bot]

Re: Limit Exercise

Take out $\displaystyle \sqrt{x}$ from the denominator of the result you did arrive at.

Hello Daon!

Thanks for your comment.

DOH!

Yes, I realized how simple it is after spending a few more minutes with the exercise during my coffee break this morning.

I feel somewhat sheepish that I did not think of factoring out an x from (x + a).

I'm not sure why, but I seem to be intimidated by radicals! Silly.

My first attempts at getting the (1 + a/x) form also failed because I have tunnel vision.

After I multiplied and divided the original expression by its conjugate, I went to multiply by sqrt(x^2)/sqrt(x^2). If I had written x/sqrt(x^2) instead, then I believe that I would have gone straight to the form posted by PKA.

Why did I not do this? Because my tunnel vision told me that a fraction is equal to one when the numerator LOOKS the same as the denominator, as opposed to HAVING the same value.

(Good grief.)

On the positive side, the guidance that I received from the contributors to this discussion has expanded my vision.

Cheers,

~ Mark

 

[Deleted member 4993]

Re: Limit Exercise

- No it is indeterminate --> 0 times infinity

Hi Subhotosh:

Would you say that the following statement is correct?

"When multiplying a 'number' with infinite magnitude by zero, the result is unknown."

In other words, we say that the Zero-Product Property of Real Numbers does not apply when infinity is involved.

Cheers,

~ Mark

There is a good discussion at the following site:

http://www.physicsforums.com/showthread.php?t=118941

In a nutshell:

if we have

f(x) = 1/x

and

g(x) = nx

then

lim(x -> inf) [ f(x) . g(x)] = 0 * inf.

but

lim(x -> inf) [ f(x) . g(x)] = 0 * inf. = 1/x * n*x = n

That limit will change with the value of 'n' - hence

0 * inf = n --> which can be any value like 0/0