$[-/1$ Points$]$

Evaluate the function $f(x)$ at the given numbers $($correct to six decimal places$).$

$f(x)=\frac{x^2-5x}{x^2-4x-5},x=0,-0\mathrm{.}5,-0\mathrm{.}9,-0\mathrm{.}95,-0\mathrm{.}99,-0\mathrm{.}999,-2,-1\mathrm{.}5,-1\mathrm{.}1,-1\mathrm{.}01,-1\mathrm{.}001$

$f(0)=q,$

$f(-0\mathrm{.}5)=q,$

$f(-0\mathrm{.}9)=$

$f(-0\mathrm{.}95)=$

$f(-0\mathrm{.}99)=$

$f(-0\mathrm{.}999)=$

$f(-2)=q,$

$f(-1\mathrm{.}5)=$

$f(-1\mathrm{.}1)=$

$f(-1\mathrm{.}01)=$

$f(-1\mathrm{.}001)=q,$

Guess the value of the limit of $f(x)$ as $x$ approaches $-1,$ correct to six decimal places. $($If an answer does not exist, enter DNE.$)$

$\underset{x-1}{lim}\frac{x^2-5x}{x^2-4x-5}=$

$q,$

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