Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

\[\sqrt{x^2-3x}-x \cdot \frac{\sqrt{x^2-3x}+x}{\sqrt{x^2-3x}+x}\]\[=\frac{(x^2-3x)+x\sqrt{x^2-3x}-x\sqrt{x^2-3x}-x^2}{\sqrt{x^2-3x}+x}\]\[=\frac{-3x}{\sqrt{x^2-3x}+x}\]

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

Ah, =_= ok.

Answer 9

What if we... divided by the smallest power. \[\large \lim_{x\rightarrow \infty} \frac{-\dfrac{3x}{x}}{\sqrt{\dfrac{x^2}{x} - \dfrac{3x}{x}} +\dfrac{x}{x}}\]

Answer 10

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

Answer 11

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.

Answer 12

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

Answer 13

Oh,I was forgetting a fundamental rule... \[\sqrt{x^2} = |x| = x ~~~\text{as x goes towards infinity}\]Therefore

\[\large \lim_{x\rightarrow \infty} \frac{-\dfrac{3x}{x}}{\sqrt{\dfrac{x^2}{x^2} - \dfrac{3x}{x^2}} +\dfrac{x}{x}}\]\[\large \lim_{\rightarrow \infty}\frac{-3}{\sqrt{1-\dfrac{3}{x}}+1}\]\[\large \frac{-3}{\sqrt{1}+1} = -\frac{3}{2}\]

Answer 14

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

Answer 15

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....

Answer 16

Hmm.. I think that makes sense

Answer 17

Oh wow, typos galore in my latex format in the above post. smh. I hope you understood what I wrote up there LOL

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

\[x\sqrt{1-\frac{3}{x}}-x = \textstyle x[ 1 + \frac{1}{2}(-\frac{3}{x}) - \frac{1}{8}(\frac{3}{x})^2 +...]-x = ...\]

and there's your -3/2