Section 2.3 : One-Sided Limits
2. 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 = - 2
- a = 1
- a = 3
- a = 5

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a a = - 2 Show SolutionFrom 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}}}}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, 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}}c 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}}}}d 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}}