Section 2.3 : One-Sided Limits
1. Below is the graph of f(x). For each of the given points determine the value of f(a), lim, \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 a = - 4 Show SolutionFrom 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}}}}b 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}}c 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}}}}d 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.