Question

Lim. f(x)=(x^2-x)/(2x^2 +5x-7) x → 1 Someone can help me ?

Answer 1

See below.

Explanation:

`(x^2-x)/(2x^2 +5x-7) = (x(x-1))/(2(x+7/2)(x-1)) = x/(2(x+7/2))`

then

`lim_(x->1)(x^2-x)/(2x^2 +5x-7) = lim_(x->1)x/(2(x+7/2)) = 1/9`

Answer 2

`lim_(xrarr1) (x^2-x)/(2x^2+5x-7)=color(red)(1/9)`

Explanation:

Note that in a case like the the object is always to reduce the numerator an denominator so that substitution of the limit value (in this case `1`) does not result in the undefined form: `0/0`.

After factoring
`color(white)("XXX")x^2-x=x(x-1)`
and
`color(white)("XXX")2x^2+5x-7=(2x+7)(x-1)`

Provided `x` is not quite equal to `1` (so that `x-1!=0`)
`lim_(xrarr1) (x^2-x)/(2x^2+5x-7)`

`color(white)("XXX")=lim_(xrarr1) (xcancel(""(x-1)))/((2x+7)cancel(""(x-1)))`

`color(white)("XXX")=1/(2 * 1 +7)`

`color(white)("XXX")=1/9`