- CoconutJJ

LIM{x -> inf} [sqrt( (X^2) - 3x) - x]

- schrodinger

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- CoconutJJ

\[\lim_{x \rightarrow \infty}(\sqrt{x^2-3x}-x)\]

- Jhannybean

\[\large \lim_{x\rightarrow \infty} \sqrt{x^2-3x}-x\] simplify this by multiplying it by its conjugate?

- CoconutJJ

I tried that, but then the denominator becomes a problem... :(

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## More answers

- 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

Take the derivative of \(\sqrt{x^2-3x}+x\) and see what you get first.

- Jhannybean

This limit should come out to \(-\dfrac{3}{2}\)

- CoconutJJ

yea.. I can use l'hopital's rule, however I'm only on chapter 1 and the teacher won't allow that

- Jhannybean

Ah, =_= ok.

- 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

however, when you move x into the radical it becomes x^2

- 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

Hmm I can't seem to find the correct answer with substitution either.. I just end up with the same equation again

- 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

\(\lim_{x \rightarrow \infty}(\sqrt{x^2-3x}-x)\)
\(=\lim_{x \rightarrow \infty}(x\sqrt{1-\frac{3}{x}}-x)\)
then expand that \(sqrt\) out via binomial

- Jhannybean

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

Hmm.. I think that makes sense

- 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

So basically dividing by \(x\) OUTSIDE the square root \(\equiv\) to dividing by \(x^2\) INSIDE the square root.

- CoconutJJ

hmmm but how did you (+x/x) I thought it was (-x)

- Jhannybean

What do you mean by (+x/x)?

- 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

|dw:1442968334574:dw|

- 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

oh so we can use both positive and negative.. I guess..?

- 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

doesn't \[\sqrt{x^2-3x}\] as x-> infinity tend to positive infinity though?

- 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

hmm.. ok THANK YOU!

- Jhannybean

Do you understand it fully?

- 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

hint: think about how we came to simplify the original function.

- CoconutJJ

|dw:1442969305759:dw|

- Jhannybean

original function: \(\color{red}{\sqrt{x^2+3x}-x}\)
simplified function: \(\color{blue}{\dfrac{-3x}{\sqrt{x^2-3x}+x}}\)

- Jhannybean

somewhere in YOUR simplification process you changed the \(+x\) to a \(-x\)

- CoconutJJ

ohhhhhhhhh..... I seeee... sorry.

- Jhannybean

No worries, lol

- CoconutJJ

okay.. now it makes sense

- CoconutJJ

THANKS, once again. :)

- 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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