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Section 2.2 : The Limit

6. Below is the graph of \(f\left( x \right)\). For each of the given points determine the value of \(f\left( a \right)\) and \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\). If any of the quantities do not exist clearly explain why.

  1. \(a = - 2\)
  2. \(a = - 1\)
  3. \(a = 1\)
  4. \(a = 3\)
TheLimit_Ex6

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a \(a = - 2\) Show Solution

Because there is no closed dot for \(x = - 2\) we can see that,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{f\left( { - 2} \right)\,\,\,{\mbox{does not exist}}}}\]

We can also see that as we approach \(x = - 2\) from both sides the graph is not approaching a value from either side and so we get,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to - 2} f\left( x \right)\,\,\,{\mbox{does not exist}}}}\]

b \(a = - 1\) Show Solution

From the graph we can see that,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{f\left( { - 1} \right) = 3}}\]

because the closed dot is at the value of \(y = 3\).

We can also see that as we approach \(x = - 1\) from both sides the graph is approaching the same value, 1, and so we get,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to - 1} f\left( x \right) = 1}}\]

Always recall that the value of a limit does not actually depend upon the value of the function at the point in question. The value of a limit only depends on the values of the function around the point in question. Often the two will be different.


c \(a = 1\) Show Solution

Because there is no closed dot for \(x = 1\) we can see that,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{f\left( 1 \right)\,\,\,{\mbox{does not exist}}}}\]

We can also see that as we approach \(x = 1\) from both sides the graph is approaching the same value, -3, and so we get,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to 1} f\left( x \right) = - 3}}\]

Always recall that the value of a limit does not actually depend upon the value of the function at the point in question. The value of a limit only depends on the values of the function around the point in question. Therefore, even though the function doesn’t exist at this point the limit can still have a value.


d \(a = 3\) Show Solution

From the graph we can see that,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{f\left( 3 \right) = 4}}\]

because the closed dot is at the value of \(y = 4\).

We can also see that as we approach \(x = 3\) from both sides the graph is approaching the same value, 4, and so we get,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to 3} f\left( x \right) = 4}}\]