what's the limit of [(3(x^2) +6)^(1/2)]/[5-2x] if the x goes to negative infinity?

\[\lim_{x \rightarrow \infty} {(3x^2 + 6)^{1/2} \over 5-2x}\] For this I would probably recognize it as something in the form of \(\frac{\infty}{\infty}\) and fall back on l'Hopital's rule.

I agree with polpak.

so would the answer be infinity?

I don't think so, no.

sorry im lost.... how do you find the answer? i was going to divide both numerator and denominator by x but then i got stuck with how to deal with the sqrt....

No, with l'Hopital's rule the limit is the limit of the ratio of the derivative of the numerator and the denominator

You can't do l'hospital's. Its an infinite loop.

http://www.wolframalpha.com/input/?i=lim+of+(3x^2%2B6)^(1%2F2)%2F(5-2x)%2Cx%2C-infinity

I see that now that I'm trying it ;p

Annoying.

Agreed.

Ah.. I guess the best thing to do is just factor out and cancel the x. \[\lim_{x \rightarrow \infty} {(3x^2 + 6)^{1/2} \over 5-2x}\]\[=\lim_{x \rightarrow \infty} {x(3 + \frac{6}{x^2})^{1/2} \over x(-2+\frac{5}{x})}\]\[=\lim_{x \rightarrow \infty} {(3 + \frac{6}{x^2})^{1/2} \over -2+\frac{5}{x}} = {\sqrt{3}\over -2}\]

Oh, you said -infinity. In that case it's positive because

-2 * -1 = 2.

is it okay to divide out the x within the sqrt....?

I factored an \(x^2\) from each term then took the square root to get just an x to cancel with the one on bottom.

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