# What is the limit of  (x^2 − 5x)/(x^2 − 4x − 5) as x approaches 5?

May 25, 2018

$\frac{5}{6}$

#### Explanation:

${x}^{2} - 5 x = x \left(x - 5\right)$
and
${x}^{2} - 4 x - 5 = \left(x - 5\right) \left(x + 1\right)$ thus we get
$\frac{x \cdot \left(x - 5\right)}{\left(x - 5\right) \left(x + 1\right)} = \frac{x}{x + 1}$ for $x \ne 5$

May 25, 2018

lim_(xrarr5)(x^2 − 5x)/(x^2 − 4x − 5)=lim_(xrarr5)[2x-5]/[2x-4]=[10-5]/[10-4]=5/6

#### Explanation:

lim_(xrarr5)(x^2 − 5x)/(x^2 − 4x − 5)=[25-25]/[25-25]=0/0

$\text{L'hospital Rule}$

since the direct compensation product equal $\frac{0}{0}$ we will use L'hospital Rule
$\textcolor{red}{{\lim}_{t \rightarrow a} \frac{f ' \left(x\right)}{g \left(' x\right)}}$

f(x)=x^2 − 5x

$f ' \left(x\right) 2 x - 5$

g(x)=x^2 − 4x − 5

$g ' \left(x\right) = 2 x - 4$

lim_(xrarr5)(x^2 − 5x)/(x^2 − 4x − 5)=lim_(xrarr5)[2x-5]/[2x-4]=[10-5]/[10-4]=5/6

May 25, 2018

$\setminus \frac{5}{6}$

#### Explanation:

You can factor $x$ in the numerator:

${x}^{2} - 5 x = x \left(x - 5\right)$

You can factor the denominator by finding the roots of the polynomial. In this case, we can use the sum/product method: if the coefficient of ${x}^{2}$ is $1$, then we can write the equation as ${x}^{2} - s x + p$, where $s$ is the sum of the solutions and $p$ is their product.

So, we're looking for two numbers ${x}_{1}$ and ${x}_{2}$ such that:

${x}_{1} + {x}_{2} = 4$
${x}_{1} {x}_{2} = - 5$

So, ${x}_{1} = - 1$ and ${x}_{2} = 5$

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

The fraction becomes

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

So, when $x$ approaches $5$, the limit is $\setminus \frac{5}{6}$