A community for students.
Here's the question you clicked on:
 0 viewing
CoconutJJ
 one year ago
LIM{x > inf} [sqrt( (X^2)  3x)  x]
CoconutJJ
 one year ago
LIM{x > inf} [sqrt( (X^2)  3x)  x]

This Question is Closed

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \infty}(\sqrt{x^23x}x)\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3\[\large \lim_{x\rightarrow \infty} \sqrt{x^23x}x\] simplify this by multiplying it by its conjugate?

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0I tried that, but then the denominator becomes a problem... :(

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3\[\sqrt{x^23x}x \cdot \frac{\sqrt{x^23x}+x}{\sqrt{x^23x}+x}\]\[=\frac{(x^23x)+x\sqrt{x^23x}x\sqrt{x^23x}x^2}{\sqrt{x^23x}+x}\]\[=\frac{3x}{\sqrt{x^23x}+x}\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Take the derivative of \(\sqrt{x^23x}+x\) and see what you get first.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3This limit should come out to \(\dfrac{3}{2}\)

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0yea.. I can use l'hopital's rule, however I'm only on chapter 1 and the teacher won't allow that

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3What if we... divided by the smallest power. \[\large \lim_{x\rightarrow \infty} \frac{\dfrac{3x}{x}}{\sqrt{\dfrac{x^2}{x}  \dfrac{3x}{x}} +\dfrac{x}{x}}\]

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0however, when you move x into the radical it becomes x^2

Empty
 one year ago
Best ResponseYou've already chosen the best response.0substitute in: \[x^23x=y^2\] solve for x to plug it in everywhere and you get: \[\lim_{y \to \infty} \sqrt{y^2}\frac{3+\sqrt{9+4y^2}}{2}\] \[\frac{3}{2}+\lim_{y \to \infty}y\frac{\sqrt{9+4y^2}}{2} \] Not really sure if this is right path to go or not, but we're getting that 3/2 that the great wolfram alpha gives.

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0Hmm I can't seem to find the correct answer with substitution either.. I just end up with the same equation again

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Oh,I was forgetting a fundamental rule... \[\sqrt{x^2} = x = x ~~~\text{as x goes towards infinity}\]Therefore \[\large \lim_{x\rightarrow \infty} \frac{\dfrac{3x}{x}}{\sqrt{\dfrac{x^2}{x^2}  \dfrac{3x}{x^2}} +\dfrac{x}{x}}\]\[\large \lim_{\rightarrow \infty}\frac{3}{\sqrt{1\dfrac{3}{x}}+1}\]\[\large \frac{3}{\sqrt{1}+1} = \frac{3}{2}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0\(\lim_{x \rightarrow \infty}(\sqrt{x^23x}x)\) \(=\lim_{x \rightarrow \infty}(x\sqrt{1\frac{3}{x}}x)\) then expand that \(sqrt\) out via binomial

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3When you're simplifying (idk if that's the right word for it) outside of the square root... you have to observe where x is tending to. Since x is moving towards \(\infty\) we can divide out the terms by \(x\) as opposed to \(x\) if \(x\rightarrow \infty\) . Does that make sense....

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0Hmm.. I think that makes sense

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Oh wow, typos galore in my latex format in the above post. smh. I hope you understood what I wrote up there LOL

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3So basically dividing by \(x\) OUTSIDE the square root \(\equiv\) to dividing by \(x^2\) INSIDE the square root.

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0hmmm but how did you (+x/x) I thought it was (x)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3What do you mean by (+x/x)?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0\[x\sqrt{1\frac{3}{x}}x = \textstyle x[ 1 + \frac{1}{2}(\frac{3}{x})  \frac{1}{8}(\frac{3}{x})^2 +...]x = ...\] and there's your 3/2

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442968334574:dw

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Basically if your was \(\lim_{x\rightarrow \infty}\) then we would have to take the equivalent of x\(^2\) outside the square root, which is \(x\) /

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0oh so we can use both positive and negative.. I guess..?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3\[\color{red}{\sqrt{x^2} = x = \pm x}\] This portion is what you have to pay attention to. This little tid bit lets you know that the simplification of your function everywhere outside the square is dependant on whether your function tends to \(+\infty\) or \(\infty\)

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0doesn't \[\sqrt{x^23x}\] as x> infinity tend to positive infinity though?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3yes it does, but see how your variables are under the square root sign? that means the variable with the highest power under the square root is x\(^2\), therefore you're going to divide both terms by x\(^2\)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Do you understand it fully?

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0yea.. I guess but I'm still confused on how (x) turned to (+x/x) shown in the drawing. didn't you divide every term by +x not x ?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3hint: think about how we came to simplify the original function.

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442969305759:dw

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3original function: \(\color{red}{\sqrt{x^2+3x}x}\) simplified function: \(\color{blue}{\dfrac{3x}{\sqrt{x^23x}+x}}\)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3somewhere in YOUR simplification process you changed the \(+x\) to a \(x\)

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0ohhhhhhhhh..... I seeee... sorry.

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0okay.. now it makes sense

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0THANKS, once again. :)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3\[\large \lim_{x\rightarrow \infty} \sqrt{x^23x}x\]\[\large \lim_{x\rightarrow \infty} \sqrt{x^23x}x \cdot \frac{\sqrt{x^23x}+x}{\sqrt{x^23x}+x}\]\[\large \lim_{x\rightarrow \infty} \frac{3x}{\sqrt{x^23x}+x}\]\[\large \lim_{x\rightarrow \infty} \frac{\dfrac{3x}{x}}{\sqrt{\dfrac{x^2}{x^2}  \dfrac{3x}{x^2}} +\dfrac{x}{x}}\]\[\large \lim_{x\rightarrow \infty}\frac{3}{\sqrt{1\dfrac{3}{x}}+1}\]\[=\boxed{\frac{3}{2}}\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.