Question

Question #a3430

Answer 1

`1/2`

Explanation:

I am assuming the expression should be this, because the terms weren't collected in the expression given.

For: `(2x^2-x+1)/(4x^2-3x-1)`

To simplify this, notice that the terms of `x^2` change much more rapidly than the terms of `x`. This means we can just concentrate on the `x^2` terms.

So we have:

`(2x^2)/(4x^2)`

Cancelling like factors:

`(2cancel(x^2))/(4cancel(x^2))=2/4=1/2`

So:

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

You may feel that somehow we have avoided taking the limit. If you go back to:

`(2x^2)/(4x^2)= (1x^2)/(2x^2)`

Notice that `x^2` in both numerator and denominator are increasing at exactly the same rate, so it would be reasonable to suggest they cancel each other out.

Hope this helps you.