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July 23, 2014

July 23, 2014

Posted by **joc** on Tuesday, December 7, 2010 at 10:00am.

- calculus -
**MathMate**, Tuesday, December 7, 2010 at 10:13amI assume you mean

radical (9x²+x)-3x

Multiply top and bottom by the conjugate

[√(9x²+x)-3x]

=[√(9x²+x)-3x]*[√(9x²+x)+3x] / √[(9x²+x)+3x]

=(9x²+x-9x²)/[√(9x²+x)+3x]

=x/[√(9x²+x)+3x]

Now evaluate the limit as x->∞

Lim x/[√(9x²+x)+3x]

= x / 6x

= 1/6

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