CoconutJJ
  • CoconutJJ
LIM{x -> inf} [sqrt( (X^2) - 3x) - x]
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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CoconutJJ
  • CoconutJJ
\[\lim_{x \rightarrow \infty}(\sqrt{x^2-3x}-x)\]
Jhannybean
  • Jhannybean
\[\large \lim_{x\rightarrow \infty} \sqrt{x^2-3x}-x\] simplify this by multiplying it by its conjugate?
CoconutJJ
  • CoconutJJ
I tried that, but then the denominator becomes a problem... :(

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Jhannybean
  • Jhannybean
\[\sqrt{x^2-3x}-x \cdot \frac{\sqrt{x^2-3x}+x}{\sqrt{x^2-3x}+x}\]\[=\frac{(x^2-3x)+x\sqrt{x^2-3x}-x\sqrt{x^2-3x}-x^2}{\sqrt{x^2-3x}+x}\]\[=\frac{-3x}{\sqrt{x^2-3x}+x}\]
Jhannybean
  • Jhannybean
Take the derivative of \(\sqrt{x^2-3x}+x\) and see what you get first.
Jhannybean
  • Jhannybean
This limit should come out to \(-\dfrac{3}{2}\)
CoconutJJ
  • CoconutJJ
yea.. I can use l'hopital's rule, however I'm only on chapter 1 and the teacher won't allow that
Jhannybean
  • Jhannybean
Ah, =_= ok.
Jhannybean
  • Jhannybean
What 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
  • CoconutJJ
however, when you move x into the radical it becomes x^2
Empty
  • Empty
substitute in: \[x^2-3x=y^2\] 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
  • CoconutJJ
Hmm I can't seem to find the correct answer with substitution either.. I just end up with the same equation again
Jhannybean
  • Jhannybean
Oh,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
  • IrishBoy123
\(\lim_{x \rightarrow \infty}(\sqrt{x^2-3x}-x)\) \(=\lim_{x \rightarrow \infty}(x\sqrt{1-\frac{3}{x}}-x)\) then expand that \(sqrt\) out via binomial
Jhannybean
  • Jhannybean
When 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
  • CoconutJJ
Hmm.. I think that makes sense
Jhannybean
  • Jhannybean
Oh wow, typos galore in my latex format in the above post. smh. I hope you understood what I wrote up there LOL
Jhannybean
  • Jhannybean
So basically dividing by \(x\) OUTSIDE the square root \(\equiv\) to dividing by \(x^2\) INSIDE the square root.
CoconutJJ
  • CoconutJJ
hmmm but how did you (+x/x) I thought it was (-x)
Jhannybean
  • Jhannybean
What do you mean by (+x/x)?
IrishBoy123
  • IrishBoy123
\[x\sqrt{1-\frac{3}{x}}-x = \textstyle x[ 1 + \frac{1}{2}(-\frac{3}{x}) - \frac{1}{8}(\frac{3}{x})^2 +...]-x = ...\] and there's your -3/2
CoconutJJ
  • CoconutJJ
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Jhannybean
  • Jhannybean
Basically 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
  • CoconutJJ
oh so we can use both positive and negative.. I guess..?
Jhannybean
  • Jhannybean
\[\color{red}{\sqrt{x^2} = |x| = \pm x}\] 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
  • CoconutJJ
doesn't \[\sqrt{x^2-3x}\] as x-> infinity tend to positive infinity though?
Jhannybean
  • Jhannybean
yes 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\(^2\)
CoconutJJ
  • CoconutJJ
hmm.. ok THANK YOU!
Jhannybean
  • Jhannybean
Do you understand it fully?
CoconutJJ
  • CoconutJJ
yea.. 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
  • Jhannybean
hint: think about how we came to simplify the original function.
CoconutJJ
  • CoconutJJ
|dw:1442969305759:dw|
Jhannybean
  • Jhannybean
original function: \(\color{red}{\sqrt{x^2+3x}-x}\) simplified function: \(\color{blue}{\dfrac{-3x}{\sqrt{x^2-3x}+x}}\)
Jhannybean
  • Jhannybean
somewhere in YOUR simplification process you changed the \(+x\) to a \(-x\)
CoconutJJ
  • CoconutJJ
ohhhhhhhhh..... I seeee... sorry.
Jhannybean
  • Jhannybean
No worries, lol
CoconutJJ
  • CoconutJJ
okay.. now it makes sense
CoconutJJ
  • CoconutJJ
THANKS, once again. :)
Jhannybean
  • Jhannybean
\[\large \lim_{x\rightarrow \infty} \sqrt{x^2-3x}-x\]\[\large \lim_{x\rightarrow \infty} \sqrt{x^2-3x}-x \cdot \frac{\sqrt{x^2-3x}+x}{\sqrt{x^2-3x}+x}\]\[\large \lim_{x\rightarrow \infty} \frac{-3x}{\sqrt{x^2-3x}+x}\]\[\large \lim_{x\rightarrow \infty} \frac{-\dfrac{3x}{x}}{\sqrt{\dfrac{x^2}{x^2} - \dfrac{3x}{x^2}} +\dfrac{x}{x}}\]\[\large \lim_{x\rightarrow \infty}\frac{-3}{\sqrt{1-\dfrac{3}{x}}+1}\]\[=\boxed{-\frac{3}{2}}\]

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