Limit of (sqrt(3x^2+8x+6)-sqrt(3x^2+3x+1)) as x -> infty

NYC

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Oct 20, 2005
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Hey, having some trouble.
The limit as X approaches infinity of (sqrt(3x^2+8x+6)-sqrt(3x^2+3x+1))

Thinking about the problem, it seems that X would approach 0 in an infinity minus infinity case. However, my teacher has taught that an infinity minus infinity case is never equal to zero, even in comparable forms of infinity (for example, x^2 minus X^2 as X approaches infinity would not equal 0). Rather, he says that the problem needs to be rephrased.

I have tried multiplying (1/X)/(1/X) through (using 1 as the denominator for both terms originally in the limit), and while that works to get a limit for the top of the fraction, the 1/X in the denominator approaches zero and seems to have no way to be dealt with.

Any help is appreciated--thanks!
 
[sqrt(3x^2+8x+6)-sqrt(3x^2+3x+1)]*[sqrt(3x^2+8x+6)+sqrt(3x^2+3x+1)]/[sqrt(3x^2+8x+6)+sqrt(3x^2+3x+1)]

[(3x^2+8x+6)-(3x^2+3x+1)]/[sqrt(3x^2+8x+6)+sqrt(3x^2+3x+1)]

(5x + 5)/[sqrt(3x^2+8x+6)+sqrt(3x^2+3x+1)]

sqrt(25x^2+50x+25)/[sqrt(3x^2+8x+6)+sqrt(3x^2+3x+1)]

divide all terms by x^2 ...

sqrt(25 + 50/x + 25/x^2)/[sqrt(3 + 8/x + 6/x^2) + sqrt(3 + 3/x + 1/x^2)]

as x -> infinity, numerator approaches sqrt(25) = 5, denominator approaches
sqrt(3) + sqrt(3) = 2sqrt(3)

limit is 5/[2sqrt(3)].
 
Many thanks! I had actually considered doing the conjugate, but it looked like it would have turned out messy.
 
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