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### Section 2.3 : One-Sided Limits

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.

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

a $$a = - 4$$ 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,

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

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.