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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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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:
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.
Posted by giorgiana1976 on May 11, 2011 at 2:51 PM (Answer #1)
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
Posted by justaguide on May 11, 2011 at 2:57 PM (Answer #2)
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