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Evaluate the limit of the function y if x goes to infinite? y=(3x^2-4x+1)/(-8x^2+5)

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sensitiv | Student | eNoter

Posted May 11, 2011 at 2:45 PM via web

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Evaluate the limit of the function y if x goes to infinite?

y=(3x^2-4x+1)/(-8x^2+5)

Tagged with calculus, function, infinite, limit, math

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giorgiana1976 | College Teacher | Valedictorian

Posted May 11, 2011 at 2:51 PM (Answer #1)

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Since the values of x approach infinite, we'll calculate the limit by factorizing both, numerator and denominator, by x^2, to create strings whose limits is zero:

lim (3x^2-4x+1)/(-8x^2+5) = lim x^2(3-5/x+1/x^2)/x^2(-8+5/x^2)

We'll reduce both, numerator and denominator, by x^2:

lim (3-5/x+1/x^2)/(-8+5/x^2)

We'll replace x by infinite:

lim (3-5/x+1/x^2)/(-8+5/x^2) = [lim3-lim (5/x)+ lim(1/x^2)]/[lim(-8) + lim(5/x^2)] = (3 - 5/infinite + 1/infinite)/(-8 + 5/infinite)

lim (3x^2-4x+1)/(-8x^2+5) = (3-0+0)/(-8+0)

lim (3x^2-4x+1)/(-8x^2+5) = -3/8

We notice that the limit is the ratio of leding coefficients of numerator and denominator.

The requested limit of the function is: lim (3x^2-4x+1)/(-8x^2+5) = -3/8.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 11, 2011 at 2:57 PM (Answer #2)

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We have to find lim x--> inf. [(3x^2-4x+1)/(-8x^2+5)]

substituting x = inf., gives the indeterminate form inf./inf., we can use l'Hopital's rule and substitute the numerator and denominator with their derivatives.

=> lim x--> inf. [(6x - 4)/(-16x)]

=> lim x--> inf. [(3x - 2)/(-8x)]

Again apply l'Hopital's rule as x = inf. gives the indeterminate form inf./inf.

=> lim x--> inf. [(3/-8]

The value of the limit is -3/8

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