Calculus I - One-Sided Limits

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Section 2.3 : One-Sided Limits

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

$a = - 2$
$a = 1$
$a = 3$
$a = 5$

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$a = - 2$ Show Solution

From the graph we can see that,

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

because the closed dot is at the value of $y = - 1$ .

We can also see that as we approach $x = - 2$ from the left the graph is not approaching a single value, but instead oscillating wildly, and as we approach from the right the graph is approaching a value of -1. Therefore, we get,

$\require{bbox} \bbox[4pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to \, - {2^{\, - }}} f\left( x \right){\mbox{ does not exist}}\hspace{0.25in} \& \hspace{0.25in},\mathop {\lim }\limits_{x \to \, - {2^{\, + }}} f\left( x \right) = - 1}}$

Recall that in order for limit to exist the function must be approaching a single value and so, in this case, because the graph to the left of $x = - 2$ is not approaching a single value the left-hand limit will not exist. This does not mean that the right-hand limit will not exist. In this case the graph to the right of $x = - 2$ is approaching a single value the right-hand limit will exist.

Now, because the two one-sided limits are different we know that,

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

$a = 1$ Show Solution

From the graph we can see that,

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

because the closed dot is at the value of $y = 4$ .

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,

$\require{bbox} \bbox[4pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to \,{1^{\, - }}} f\left( x \right) = 3\hspace{0.25in}\& \hspace{0.25in}\mathop {\lim }\limits_{x \to \,{1^{\, + }}} f\left( x \right) = 3}}$

The two one-sided limits are the same so we know,

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

$a = 3$ Show Solution

From the graph we can see that,

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

because the closed dot is at the value of $y = - 2$ .

We can also see that as we approach $x = 2$ from the left the graph is approaching a value of 1 and as we approach from the right the graph is approaching a value of -3. Therefore, we get,

$\require{bbox} \bbox[4pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to \,{3^{\, - }}} f\left( x \right) = 1\hspace{0.25in}\& \hspace{0.25in}\mathop {\lim }\limits_{x \to \,{3^{\, + }}} f\left( x \right) = - 3}}$

Now, because the two one-sided limits are different we know that,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to \,3} f\left( x \right)\,\,{\mbox{does not exist}}}}$

$a = 5$ Show Solution

From the graph we can see that,

$\require{bbox} \bbox[2pt,border:1px solid black]{{f\left( 5 \right) = 4}}$

because the closed dot is at the value of $y = 4$ .

We can also see that as we approach $x = 5$ from both sides the graph is approaching the same value, 4, and so we get,

$\require{bbox} \bbox[4pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to \,{5^{\, - }}} f\left( x \right) = 4\hspace{0.25in}\& \hspace{0.25in}\mathop {\lim }\limits_{x \to \,{5^{\, + }}} f\left( x \right) = 4}}$

The two one-sided limits are the same so we know,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to \,5} f\left( x \right) = 4}}$