Question

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

Answer 1

\[\lim_{x \rightarrow \infty} {(3x^2 + 6)^{1/2} \over 5-2x}\]

For this I would probably recognize it as something in the form of \(\frac{\infty}{\infty}\) and fall back on l'Hopital's rule.

Answer 2

I agree with polpak.

Answer 3

so would the answer be infinity?

Answer 4

I don't think so, no.

Answer 5

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....

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

Annoying.

Answer 11

Agreed.

Answer 12

Ah.. I guess the best thing to do is just factor out and cancel the x.

\[\lim_{x \rightarrow \infty} {(3x^2 + 6)^{1/2} \over 5-2x}\]\[=\lim_{x \rightarrow \infty} {x(3 + \frac{6}{x^2})^{1/2} \over x(-2+\frac{5}{x})}\]\[=\lim_{x \rightarrow \infty} {(3 + \frac{6}{x^2})^{1/2} \over -2+\frac{5}{x}} = {\sqrt{3}\over -2}\]

Answer 13

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

Answer 14

-2 * -1 = 2.

Answer 15

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

Answer 16

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.