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Math Help - Limits of rational functions with absolute values in the denominator

  1. #1
    Member rowe's Avatar
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    Limits of rational functions with absolute values in the denominator

    Consider \lim_{x \to -6} \frac{2x+12}{|x+6|}.

    I thought the best way to approach this was to consider the two sides of the limit seperately.

    \lim_{x \to -6^+} \frac{2x+12}{|x+6|}

    = \lim_{x \to -6^+} \frac{2x+12}{6-x} because x < 0.

    Now the limit will evaluate to 0. I had to check numerically to discover the limits on both sides of -6 are -2 and 2 respectively, so there is no limit.

    I had a hunch from the fact that the limit seemed to be equal to 0.

    But how can I, without tedious checking values,

    a) Establish whether there is a limit or not
    b) The value of the limit if it exists?

    If I carried on applying algebraic rules and limit laws to this problem I would get the answer 0, and be none the wiser.
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    \frac{{2x + 12}}<br />
{{\left| {x + 6} \right|}} = \left\{ \begin{gathered}<br />
  2,\;x >  - 6 \hfill \\<br />
   - 2,\;x <  - 6 \hfill \\ <br />
\end{gathered}  \right.
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  3. #3
    Member rowe's Avatar
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    Quote Originally Posted by Plato View Post
    \frac{{2x + 12}}<br />
{{\left| {x + 6} \right|}} = \left\{ \begin{gathered}<br />
  2,\;x >  - 6 \hfill \\<br />
   - 2,\;x <  - 6 \hfill \\ <br />
\end{gathered}  \right.
    Can you explain how you came to that conclusion? I did it by plugging in values on either side, as I mentioned. But I really don't want to plug in values if there's perhaps a nicer way to evaluate this.
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    Quote Originally Posted by rowe View Post
    Can you explain how you came to that conclusion? I did it by plugging in values on either side, as I mentioned. But I really don't want to plug in values if there's perhaps a nicer way to evaluate this.
    It is truly elementary.
    Note that 2x+12=2(x+6)

    Then note that \left| {x + 6} \right| = \left\{ {\begin{array}{rl}<br />
   {(x + 6),} & {x >  - 6}  \\<br />
   { - (x + 6),} & {x <  - 6}  \\ \end{array} } \right.
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  5. #5
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    Just a bit more detail: For x> -6, the is \frac{2x+ 12}{x+ 6}= \frac{2(x+6)}{x+6} and, since x is not equal to -6, we can cancel to get 2. If x< -6, this is \frac{2x+12}{-(x+6)}= \frac{2(x+6)}{-(x+6)}= -2.

    Of course, you then use the basic property: "If f(x)= g(x) for all x except, possibly, x= 1, then \lim_{x\to a^+} f(x)= \lim_{x\to a^+} g(x) and \lim_{x\to a^-} f(x)= \lim_{x\to a^-} g(x)".
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