Calculus I - One-Sided Limits

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Section 2.3 : One-Sided Limits

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1. 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.

$a = - 4$
$a = - 1$
$a = 2$
$a = 4$

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$a = - 4$ Show Solution

From the graph we can see that,

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

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

We can also see that as we approach $x = - 4$ from the left the graph is approaching a value of 3 and as we approach from the right the graph is approaching a value of -2. Therefore, we get,

$\require{bbox} \bbox[4pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to \, - {4^{\, - }}} f\left( x \right) = 3\hspace{0.25in}\hspace{0.25in}\& \hspace{0.25in}\mathop {\lim }\limits_{x \to \, - {4^{\, + }}} f\left( x \right) = - 2}}$

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

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

$a = - 1$ 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, 4, and so we get,

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

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

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

$a = 2$ 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 approaching a value of -1 and as we approach from the right the graph is approaching a value of 5. Therefore, we get,

$\require{bbox} \bbox[4pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to \,{2^{\, - }}} f\left( x \right) = - 1\hspace{0.25in}\hspace{0.25in}\& \hspace{0.25in}\mathop {\lim }\limits_{x \to \,{2^{\, + }}} f\left( x \right) = 5}}$

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

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

$a = 4$ Show Solution

Because there is no closed dot for $x = 4$ we can see that,

$\require{bbox} \bbox[2pt,border:1px solid black]{{f\left( 4 \right)\,\,\,{\mbox{does not exist}}}}$

We can also see that as we approach $x = 4$ from both sides the graph is approaching the same value, 2, and so we get,

$\require{bbox} \bbox[4pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to \,{4^{\, - }}} f\left( x \right) = 2\hspace{0.25in}\hspace{0.25in}\& \hspace{0.25in}\mathop {\lim }\limits_{x \to \,{4^{\, + }}} f\left( x \right) = 2}}$

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

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

Always recall that the value of a limit (including one-sided limits) 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 and one-sided limits can still have a value.