Calculus I - The Limit

Section 2.2 : The Limit

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.

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

$The graph of a function on $-12 \le x \le 12$ made up of three pieces. The first piece starts at (-12,4), falls to its lowest point at an open dot at (-8,-6) and then rises to a closed dot at (-2,3). There is also a closed dot at (-8,-3) above the open dot at $x = -8$. The second piece comes down from positive infinity just to the right of $x = -2$, reaches a low point of about 1 near $x = 0$, rises to about 3 near $x = 2.5$, dips slightly and ends at an open dot at (6,2). The third piece starts at a closed dot at (6,5), rises to a peak of about 5.5 near $x = 7$ and then falls through a closed dot at (10,0) to about -4 at $x = 12$.$

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a $a = - 8$ 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.

b $a = - 2$ 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,

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

c $a = 6$ 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,

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

d $a = 10$ 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,

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