How do you find lim (2x^2-4x+1)/(3x^2+5x-6) as x->oo?

1 Answer
Steve M
Mar 2, 2017

lim_(x->oo) (2x^2-4x+1)/(3x^2+5x-6) = 2/3

Explanation:

We can manipulate the limit as follows

lim_(x->oo) (2x^2-4x+1)/(3x^2+5x-6) = lim_(x->oo) (2x^2-4x+1)/(3x^2+5x-6) *(1/x^2)/(1/x^2)

" "= lim_(x->oo) (2-4/x+1/x^2)/(3+5/x-6/x^2)

" "= 2/3

As both 1/x rarr 0 and 1/x^2 rarr 0 as x rarr oo

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