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### Section 2.3 : One-Sided Limits

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.

1. $$a = - 2$$
2. $$a = 1$$
3. $$a = 3$$
4. $$a = 5$$ Show All Solutions Hide All Solutions

a $$a = - 2$$ 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,

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

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,

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