Approaching Infinity: Solving Limits with the Conjugate Method

[curiouschris]

Hi I am trying to help my son with his math course
(disclaimer: I have done very little calculus myself)


lim x $$\rightarrow$$$$\infty$$ $$\sqrt{x^2-3x}-x$$

That was hard hope you don't mind if I don't use latex. I looked at the samples but the preview never rendered what I expected :(

We got this far using the conjugate

= (sqrt(x^2-3x)-x)(sqrt(x^2-3x)+x) / (sqrt(x^2-3x)+x)

= (-3x) / (sqrt(x^2-3x)+x)

I don't know where to take it from here. if I try to factor out the x's All I get is -(0/0) obviously wrong

Any help would be gratefully appreciated.

CC

 

[Dick]

Try factoring sqrt(x^2-3*x)=sqrt(x^2*(1-3/x))=sqrt(x^2)*sqrt(1-3/x). Any ideas where to go from there?

 

[curiouschris]

Sorry I am missing something (please remember I am trying to help my son, so anything I learned was eons ago)

I don't see how sqrt(x^2-3*x) becomes sqrt(x^2*(1-3/x))

of course once it does sqrt(x^2) = x

 

[curiouschris]

To see how far I have got I understand the beginning of example 14 on this page, but I get lost at the "factor out x" part
http://www.analyzemath.com/calculus/limits/find_limits_functions.html

CC

 

[Dick]

Sorry I am missing something (please remember I am trying to help my son, so anything I learned was eons ago)

I don't see how sqrt(x^2-3*x) becomes sqrt(x^2*(1-3/x))

of course once it does sqrt(x^2) = x

x^2-3*x=(x^2)*(1-3/x). Just multiply (x^2)*(1-3/x) back out to see why. And sqrt(x^2)=x isn't quite true. sqrt(x^2)=|x| is better. No problem with asking more questions.

 

[Dick]

To see how far I have got I understand the beginning of example 14 on this page, but I get lost at the "factor out x" part
http://www.analyzemath.com/calculus/limits/find_limits_functions.html

CC

That's pretty much the same as this problem. Let's just push through on this one and you shouldn't have any problems with example 14.

 

[Stardust*]

ACHTUNG: This post contains some wrong parts in LaTeX I am not able to correct. Please jump it...

When you use the conjugate, the objective is to eliminate all the disturbing x's floating around.
If I have understood your initial steps, the limit should look like this (after using the conjugate technique):
$$\lim_{x\rightarrow \infty}\displaystyle \frac{-3x}{\sqrt{x^2-3x}+x}$$

Extract the highest terms from the denominator as suggested:
(the lower part of the fraction is:)
$$\sqrt{x^2-3x}+x=(x^2-3x)^{1/2}+x=[x^{2}(1-\frac{3x}{x^2})]^{1/2}+x=|x|(1-\frac{3}{x})^{1/2}+x$$
$$\sqrt{x^2-3x}+x=x^{2/2}(1-\frac{3}{x})^{1/2}+x=|x|[\sqrt{1-\frac{2}{x}} +1}]$$
Now try the limit and check if things get better. I hope the passages are clear.
The signum |x| is required,isn't it?

Edit: I took the wrong exponent at the beginning. I do not understand why the edit function does not work properly with me. Really sorry, I did not want to make a mess... My next post should bring a better version of this one.

 

[Dick]

When you use the conjugate, the objective is to eliminate all the disturbing x's floating around.
If I have understood your initial steps, the limit should look like this (after using the conjugate technique):
$$\lim_{x\rightarrow \infty}\displaystyle \frac{-3x}{\sqrt{x^3-3x}+x}$$

Extract the highest terms from the denominator as suggested:
(the lower part of the fraction is:)
$$\sqrt{x^3-3x}+x=(x^3-3x)^{1/2}+x=[x^{3}(1-\frac{3x}{x^3})]^{1/2}+x=x^{3/2}(1-\frac{3x}{x^3})^{1/2}+x$$
$$]\sqrt{x^3-3x}+x=x^{3/2}(1-\frac{3}{x^2})^{1/2}+x^{3/2}x^{-1/2}=x^{3/2}[\sqrt{1-\frac{2}{x^2}} +\frac{1}{\sqrt{x}}]$$
Now try the limit and check if things get better. I hope the passages are clear.

Why did you substitute x^3 for x^2 in sqrt(x^2-3x)? I don't think this is making things clearer.

 

[Stardust*]

A correction to my previous post.
Extract the highest terms from the denominator as suggested:
(the lower part of the fraction is:)
$$\sqrt{x^2-3x}+x=(x^2-3x)^{1/2}+x=[x^{2}(1-\frac{3x}{x^2})]^{1/2}+x=|x|(1-\frac{3}{x})^{1/2}+x$$

$$\sqrt{x^2-3x}+x=x^{2/2} (1-\frac{3}{x})^{1/2}+x=|x|[\sqrt{1-\frac{3}{x}} +1}]$$

Now try the limit and check if things get better. I hope the passages are clear.
The signum |x| is required,isn't it?

 

[curiouschris]

Sorry guys had to go to bed after my last response am doing this from work shhh...

A correction to my previous post.
Extract the highest terms from the denominator as suggested:
(the lower part of the fraction is:)
$$\sqrt{x^2-3x}+x=(x^2-3x)^{1/2}+x=[x^{2}(1-\frac{3x}{x^2})]^{1/2}+x=|x|(1-\frac{3}{x})^{1/2}+x$$

$$\sqrt{x^2-3x}+x=x^{2/2} (1-\frac{3}{x})^{1/2}+x=|x|[\sqrt{1-\frac{3}{x}} +1}]$$

Now try the limit and check if things get better. I hope the passages are clear.
The signum |x| is required,isn't it?

Hmm.. the latex renderer seems to aggressively cache latex so it won't re-render it even if the terms have changed.

how do we get from...
$$x^{2/2} (1-\frac{3}{x})^{1/2}+x$$ to $$|x|[\sqrt{1-\frac{3}{x}}+1}]$$ ?

I don't get where the $$+1$$ comes from $$\sqrt{x^2}$$ becomes $$|x|$$ (thanks dick) but randomly replacing the x with 1 seems wrong

What intermediate step am I missing?

CC

 

[Char. Limit]

Sorry guys had to go to bed after my last response am doing this from work shhh...


Hmm.. the latex renderer seems to aggressively cache latex so it won't re-render it even if the terms have changed.

how do we get from...
$$x^{2/2} (1-\frac{3}{x})^{1/2}+x$$ to $$|x|[\sqrt{1-\frac{3}{x}}+1}]$$ ?

I don't get where the $$+1$$ comes from $$\sqrt{x^2}$$ becomes $$|x|$$ (thanks dick) but randomly replacing the x with 1 seems wrong

What intermediate step am I missing?

CC

Well, he went from sqrt(x^2) to |x|, then we can assume x is positive (it's heading toward positive infinity), so we can substitute |x| with x. Then we get...

$$x \sqrt{1-\frac{3}{x}} + x$$

Which we then factor out the x from both terms to get...

$$x \left(\sqrt{1 - \frac{3}{x}} + 1\right)$$

 

[curiouschris]

Well, he went from sqrt(x^2) to |x|, then we can assume x is positive (it's heading toward positive infinity), so we can substitute |x| with x. Then we get...

$$x \sqrt{1-\frac{3}{x}} + x$$

Which we then factor out the x from both terms to get...

$$x \left(\sqrt{1 - \frac{3}{x}} + 1\right)$$

Thanks that was the last part of the problem. When I get a few minutes I am going to post the complete work-through. hopefully all this effort you have gone to to help me understand, will help someone else.

The main lesson I have learn't is my sons not going to get through this course unless he scrubs up on his basic algebra. Dads not a great deal of help, its been too long for me and I no longer have the intuition that only comes with practice.

Ta
CC

 

[Char. Limit]

Good luck, and I hope your son brushes up on the algebra well!