Calculus I - The Limit

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Section 2.2 : The Limit

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5. 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.

$a = - 8$
$a = - 2$
$a = 6$
$a = 10$

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$a = - 8$ Show Solution

From the graph we can see that,

$\require{bbox} \bbox[2pt,border:1px solid black]{{f\left( { - 8} \right) = - 3}}$

because the closed dot is at the value of $y = - 3$ .

We can also see that as we approach $x = - 8$ from both sides the graph is approaching the same value, -6, and so we get,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to - 8} f\left( x \right) = - 6}}$

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.

$a = - 2$ Show Solution

From the graph we can see that,

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

because the closed dot is at the value of $y = 3$ .

We can also see that as we approach $x = - 2$ from both sides the graph is approaching different values (3 from the left and doesn’t approach any value from the right). Because of this we get,

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

$a = 6$ Show Solution

From the graph we can see that,

$\require{bbox} \bbox[2pt,border:1px solid black]{{f\left( 6 \right) = 5}}$

because the closed dot is at the value of $y = 5$ .

We can also see that as we approach $x = 6$ from both sides the graph is approaching different values (2 from the left and 5 from the right). Because of this we get,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to 6} f\left( x \right)\,\,\,{\mbox{does not exist}}}}$

$a = 10$ Show Solution

From the graph we can see that,

$\require{bbox} \bbox[2pt,border:1px solid black]{{f\left( {10} \right) = 0}}$

because the closed dot is at the value of $y = 0$ .

We can also see that as we approach $x = 10$ from both sides the graph is approaching the same value, 0, and so we get,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to 10} f\left( x \right) = 0}}$