# How do you evaluate ( 5x^9-4x^2-2x+1)/(x^7-x^6+x-1) as x approaches 1?

May 18, 2016

${\lim}_{x \to 1} \frac{5 {x}^{9} - 4 {x}^{2} - 2 x + 1}{{x}^{7} - {x}^{6} + x - 1} = \frac{35}{2}$

#### Explanation:

${\lim}_{x \to 1} \frac{5 {x}^{9} - 4 {x}^{2} - 2 x + 1}{{x}^{7} - {x}^{6} + x - 1}$

$= {\lim}_{x \to 1} \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \left(5 {x}^{8} + 5 {x}^{7} + 5 {x}^{6} + 5 {x}^{5} + 5 {x}^{4} + 5 {x}^{3} + 5 {x}^{2} + x - 1\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \left({x}^{6} + 1\right)}$

$= {\lim}_{x \to 1} \frac{5 {x}^{8} + 5 {x}^{7} + 5 {x}^{6} + 5 {x}^{5} + 5 {x}^{4} + 5 {x}^{3} + 5 {x}^{2} + x - 1}{{x}^{6} + 1}$

$= \frac{5 + 5 + 5 + 5 + 5 + 5 + 5 + 1 - 1}{1 + 1}$

$= \frac{35}{2}$

$\textcolor{w h i t e}{}$
Alternative method

Alternatively you can use L'Hôpital's rule in the form:

If the following hold:

• $f \left(x\right)$ and $g \left(x\right)$ are differentiable on an open interval containing $c$, except possibly at $c$.

• ${\lim}_{x \to c} f \left(x\right) = {\lim}_{x \to c} g \left(x\right) = 0$

• ${\lim}_{x \to c} \frac{f ' \left(x\right)}{g ' \left(x\right)}$ exists

Then:

${\lim}_{x \to c} f \frac{x}{g} \left(x\right) = {\lim}_{x \to c} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

$\textcolor{w h i t e}{}$
For our example, let:

$f \left(x\right) = 5 {x}^{9} - 4 {x}^{2} - 2 x + 1$

$g \left(x\right) = {x}^{7} - {x}^{6} + x - 1$

$c = 1$

Then:

• $f \left(x\right)$ and $g \left(x\right)$ are differentiable on an open interval containing $c = 1$ with:

$f ' \left(x\right) = 45 {x}^{8} - 8 x - 2$

$g ' \left(x\right) = 7 {x}^{6} - 6 {x}^{5} + 1$

• ${\lim}_{x \to 1} f \left(x\right) = {\lim}_{x \to 1} g \left(x\right) = 0$

• ${\lim}_{x \to 1} \frac{f ' \left(x\right)}{g ' \left(x\right)} = \frac{45 - 8 - 2}{7 - 6 + 1} = \frac{35}{2}$ exists

So:

${\lim}_{x \to 1} f \frac{x}{g} \left(x\right) = {\lim}_{x \to 1} \frac{f ' \left(x\right)}{g ' \left(x\right)} = \frac{35}{2}$