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

Mar 31, 2018

See below.

#### Explanation:

$\frac{{x}^{2} - x}{2 {x}^{2} + 5 x - 7} = \frac{x \left(x - 1\right)}{2 \left(x + \frac{7}{2}\right) \left(x - 1\right)} = \frac{x}{2 \left(x + \frac{7}{2}\right)}$

then

${\lim}_{x \to 1} \frac{{x}^{2} - x}{2 {x}^{2} + 5 x - 7} = {\lim}_{x \to 1} \frac{x}{2 \left(x + \frac{7}{2}\right)} = \frac{1}{9}$

Mar 31, 2018

#### Answer:

${\lim}_{x \rightarrow 1} \frac{{x}^{2} - x}{2 {x}^{2} + 5 x - 7} = \textcolor{red}{\frac{1}{9}}$

#### 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: $\frac{0}{0}$.

After factoring
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - x = x \left(x - 1\right)$
and
$\textcolor{w h i t e}{\text{XXX}} 2 {x}^{2} + 5 x - 7 = \left(2 x + 7\right) \left(x - 1\right)$

Provided $x$ is not quite equal to $1$ (so that $x - 1 \ne 0$)
${\lim}_{x \rightarrow 1} \frac{{x}^{2} - x}{2 {x}^{2} + 5 x - 7}$

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

$\textcolor{w h i t e}{\text{XXX}} = \frac{1}{2 \cdot 1 + 7}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{1}{9}$