Section 2.2 : The Limit
4. 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.
- \(a = - 3\)
- \(a = - 1\)
- \(a = 2\)
- \(a = 4\)
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a \(a = - 3\) Show SolutionFrom 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 different values (4 from the left and -2 from the right). Because of this we get,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to - 3} f\left( x \right)\,\,\,{\mbox{does not exist}}}}\]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.
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 \(a = 2\) 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 approaching the same value, 1, and so we get,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 1}}\]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. Therefore, even though the function doesn’t exist at this point the limit can still have a value.
d \(a = 4\) Show Solution
From the graph we can see that,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{f\left( 4 \right) = 5}}\]because the closed dot is at the value of \(y = 5\).
We can also see that as we approach \(x = 4\) from both sides the graph is approaching the same value, 5, and so we get,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to 4} f\left( x \right) = 5}}\]