Question

Question #302d0

Answer 1

`lim_(x->oo) frac{x^2-4x+1}{-4x^2+3x}=-1/4`

This is because the highest degree of the polynomial in the numerator (`2`) is the same as that of the denominator (`2`), so the end behavior is the fraction of the coefficients of those terms. Or, the formal proof using L'Hospital's rule is below.

Explanation:

`lim_(x->oo) frac{x^2-4x+1}{-4x^2+3x}`

By direct substitution, this gives `oo/-oo`, which is an indeterminate case. Use L'Hospital's rule, which states that the limit above is equal to limit as `x` approaches `oo` of the derivative of the numerator over the derivative of the denominator:
`=lim_(x->oo) frac{2x-4}{-8x+3}`

Again, this limit gives `oo/-oo`, so use L'Hospital's rule again:
`=lim_(x->oo) frac{2}{-8}`

`=-1/4`