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by **jsel** on Fri Oct 15, 2010 7:42 pm

$\lim_{x \rightarrow -\infty}\, \left(\sqrt{x^2\, +\, x\, +\, 1}\, +\, x\right)$

The best thing I could think to do here is multiple the numerator and denominator by the conjugate radical sqrt(x^2 + x + 1) - x. But that will lead to a 0 in the denominator once x is factored out. Ialso tried breaking up the quadratic under the radical into factored form, but it has no solutions. Can someone give me a hint?

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by **stapel_eliz** on Fri Oct 15, 2010 9:00 pm

jsel wrote: $\lim_{x \rightarrow -\infty}\, \left(\sqrt{x^2\, +\, x\, +\, 1}\, +\, x\right)$

The best thing I could think to do here is multiple the numerator and denominator by the conjugate radical sqrt(x^2 + x + 1) - x. But that will lead to a 0 in the denominator....

Not actually.

Since $\sqrt{x^2}\, =\, |x|$ and since $x$ will be (very, very) negative, then the square root will "simplify" to $|x|\, =\, -x$ .

Where would this lead? :wink:

by **jsel** on Sun Oct 17, 2010 4:47 pm

Got it, thanks. I forgot about that property when taking the square root of x^2

by **stapel_eliz** on Sun Oct 17, 2010 9:34 pm

jsel wrote:I forgot about that property when taking the square root of x^2

Everybody does. Expect a trick question using this property on the test. :wink:

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