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Limits Involving Absolute Values

April 8th, 2009 |

Author: Merrin

Naively trying to use the limit cancellation theorem in expressions involving absolute values may give a wrong answer.

Example 1. Find the following limit

$\mathop{\lim }\limits_{x\to 3} \cfrac{\left|x-3\right|}{x-3}$

Solution 1. The key to solving problems involving absolute values is to break the absolute value into two equations over the domain. To calculate the limit, we explore the right hand and left hand limits

${\mathop{\lim }\limits_{x\to 3^{+} } \cfrac{x-3}{x-3} =\mathop{\lim }\limits_{\varepsilon \to 0^{+} } \cfrac{3+\varepsilon -3}{3+\varepsilon -3} =\mathop{\lim }\limits_{\varepsilon \to 0^{+} } \cfrac{\varepsilon }{\varepsilon } =1} \\ {\mathop{\lim }\limits_{x\to 3^{-} } \cfrac{x-3}{-(x-3)} =\mathop{\lim }\limits_{\varepsilon \to 0^{+} } \cfrac{3-\varepsilon -3}{-((3-\varepsilon )-3)} =\mathop{\lim }\limits_{\varepsilon \to 0^{+} } \cfrac{-\varepsilon }{\varepsilon } =-1}$

Since the left hand and right hand limits do not match, the limit of f(x) does not exist at x=3. Graphing the function we see that f(x) has a jump discontinuity there.

\includegraphics[width= 5in]{../images/absxm3.pdf}

\caption{Graphing $f(x) = \cfrac{\left| x-3 \right|}{x-3}$ reveals a jump discontinuity.}

Example 2. Find for which points the limit exists using the function

$f(x)=\cfrac{\left|x-3\right|}{x-1}$

Solution 2. First break the equation up into two intervals for the absolute value.

${x\ge 3\quad \, f(x)=\cfrac{x-3}{x-1} } \\ {x<3\quad f(x)=\cfrac{3-x}{x-1} }$

For this function, the limit at x=3 exists but for the limit x=1 does not.This troublesome point is in the left hand branch of the absolute value. At all other locations the function is continuous. We can further analyze the limit at the troublesome point

$\mathop{\lim }\limits_{x\to 1} \cfrac{x-3}{x-1}$

Find the left and right hand limits

$\mathop{\lim }\limits_{x\to 1^{+} } \cfrac{x-3}{x-1} =\mathop{\lim }\limits_{\varepsilon \to 0^{+} } \cfrac{1+\varepsilon -3}{1+\varepsilon -1} =\mathop{\lim }\limits_{\varepsilon \to 0^{+} } \cfrac{-2}{\varepsilon } =-\infty$

$\mathop{\lim }\limits_{x\to 1^{+} } \cfrac{x-3}{x-1} =\mathop{\lim }\limits_{\varepsilon \to 0^{+} } \cfrac{1-\varepsilon -3}{1-\varepsilon -1} =\mathop{\lim }\limits_{\varepsilon \to 0^{+} } \cfrac{2}{\varepsilon } =\infty$

$\mathop{\lim }\limits_{x\to 1} \cfrac{x-3}{x-1} =DNE$

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Limits Involving Absolute Values

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