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

### 2 Answers | Add Yours

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.**

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**