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OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

I agree with polpak.

OpenStudy (anonymous):

so would the answer be infinity?

OpenStudy (anonymous):

I don't think so, no.

OpenStudy (anonymous):

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....

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Annoying.

OpenStudy (anonymous):

Agreed.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

-2 * -1 = 2.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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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