# How do you evaluate the limit (7x^2-4x-3)/(3x^2-4x+1) as x approaches 1?

Oct 6, 2016

${\lim}_{x \to 1} \frac{7 {x}^{2} - 4 x - 3}{3 {x}^{2} - 4 x + 1} = 5$

#### Explanation:

Factoring the numerator and denominator, we can simplify the rational expression as follows:

$\frac{7 {x}^{2} - 4 x - 3}{3 {x}^{2} - 4 x + 1} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \left(7 x + 3\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \left(3 x - 1\right)} = \frac{7 x + 3}{3 x - 1}$

with exclusion $x \ne 1$

That means that the functions $\frac{7 {x}^{2} - 4 x - 3}{3 {x}^{2} - 4 x + 1}$ and $\frac{7 x + 3}{3 x - 1}$ are identical, except that the first one has a hole at $x = 0$, while the second one does not.

So we find:

${\lim}_{x \to 1} \frac{7 {x}^{2} - 4 x - 3}{3 {x}^{2} - 4 x + 1} = {\lim}_{x \to 1} \frac{7 x + 3}{3 x - 1}$

$\textcolor{w h i t e}{{\lim}_{x \to 1} \frac{7 {x}^{2} - 4 x - 3}{3 {x}^{2} - 4 x + 1}} = \frac{7 \left(\textcolor{b l u e}{1}\right) + 3}{3 \left(\textcolor{b l u e}{1}\right) - 1}$

$\textcolor{w h i t e}{{\lim}_{x \to 1} \frac{7 {x}^{2} - 4 x - 3}{3 {x}^{2} - 4 x + 1}} = \frac{7 + 3}{3 - 1}$

$\textcolor{w h i t e}{{\lim}_{x \to 1} \frac{7 {x}^{2} - 4 x - 3}{3 {x}^{2} - 4 x + 1}} = \frac{10}{2}$

$\textcolor{w h i t e}{{\lim}_{x \to 1} \frac{7 {x}^{2} - 4 x - 3}{3 {x}^{2} - 4 x + 1}} = 5$