# Limit

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latex
7 months ago
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\documentclass{article}
\usepackage{amsmath}

\newcommand{\Lim}[1]{\raisebox{0.5ex}{{$\displaystyle \lim_{#1}\;$}}}

\begin{document}

Evaluate
$\Lim{x\to 2^-}\frac{x-1+ \lfloor 1-x \rfloor}{|x^2 -7x+10|}$
\\\\
\begin{align*}
&
=\Lim{x\to 2^-}
\frac{x-1+ \lfloor 1-x \rfloor}{|x^2 -7x+10|}
\\\\
&
=\Lim{x\to 2^-}
\frac{x-1+ \lfloor 1-x \rfloor}{|(x-5)(x-2)|}
\\\\
&
=\Lim{x\to 2^-}
\frac{x-1+ \lfloor 1-x \rfloor}{|(x-5)| |(x-2)|}
\\\\
&
=\Lim{x\to 2^-}
\frac{x-1-1}{|(x-5)| |(x-2)|}
\\\\
&
=\Lim{x\to 2^-}
\frac{x-1-1}{(-)(x-5)(-)(x-2)}
\\\\
&
=\Lim{x\to 2^-}
\frac{x-2}{(-)(x-5)(-)(x-2)}
\\\\
&
=\Lim{x\to 2^-}
\frac{1}{(-)(x-5)(-)}
\\\\
&
=
\frac{1}{(-)(-3)(-)}=-\frac{1}{3}
\end{align*}

\begin{align*}
\lfloor 1-x \rfloor =
$\begin{cases} -2, & -2 \leq 1-x < -1 \Longleftrightarrow 2 < x \leq 3 \\ -1, & -1 \leq 1 -x <0 \Longleftrightarrow 1 < x \leq 2\\ 0, & 0 \leq 1-x < 1 \Longleftrightarrow 0 < x \leq 1 \end{cases}$
\end{align*}

\begin{align*}
|x-5| =
$\begin{cases} x-5, & x-5 \geq 0 \Longleftrightarrow x \geq 5 \\ -(x-5), & x-5 < 0 \Longleftrightarrow x < 5 \end{cases}$
\end{align*}

\begin{align*}
|x-2| =
$\begin{cases} x-2, & x-2 \geq 0 \Longleftrightarrow x \geq 2 \\ -(x-2), & x-2 < 0 \Longleftrightarrow x < 2 \end{cases}$
\end{align*}

\end{document}