Calculus I - One-Sided Limits (Assignment Problems)

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Section 2.3 : One-Sided Limits

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. $a = - 5$
2. $a = - 2$
3. $a = 1$
4. $a = 4$
$This graph consists of two V shaped portions. The first portion is in the domain $-7 \le x < 1$. Its point is at (-5,7) – this is an open dot, and opens downwards. There is a closed dot at (-5,3) and at (-2,2). The graph ends on the right at an open dot at (1,-3). The second portion is in the domain $1 < x \le 6$. Its point is at (4,-2) – this is an open dot, and opens upwards. The graph starts at an open dot at (1,4) and there is a closed dot at (1,6).$
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. $a = - 1$
2. $a = 1$
3. $a = 3$
$This graph consists of four horizontal line segments. The first is in the domain $x < -1$ and is at y=4. It ends at a closed dot of (-1,4). The second is in the domain $-1 < x < \le 1$ and is at y=-2. It starts at an open dot at (-1,-2) and ends at a closed dot at (1,-2). The third is in the domain $1 < x < 3$ and is at y=3. It starts at an open dot of (1,3) and ends at an open dot of (3,3). The final segment is in the domain $x>3$ and is at y=1. It starts with a closed dot at (3,1).$
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. $a = - 3$
2. $a = - 1$
3. $a = 1$
4. $a = 2$
$This graph consists of four segments. The first is in the domain $x < -3$. This segment is a decreasing function that starts at (-4,5) and ends at an open dot at (-3,2). There is also a closed dot at (-3,5). The second segment is in the domain $-3 < x \le -1$. It starts with an open dot at (-3,2) and increases until approximately (-2.2, 4.8) and then decreases until it ends at a closed dot at (-1,4). The third segment is in the domain $-1 < x < \le 2$. It starts with a closed dot at (-1,-3) has an open dot at (1,-1) and ends at a closed dot at (2,0). The final segment is in the domain $x > 2$. The graph in the segment is an oscillating function that oscillates faster and faster as it approaches x = 2 from the right and the oscillation slows down as it moves away from x=2.$
Sketch a graph of a function that satisfies each of the following conditions. \[\mathop {\lim }\limits_{x \to {1^{\, - }}} f\left( x \right) = - 2\hspace{0.75in}\mathop {\lim }\limits_{x \to {1^{\, + }}} f\left( x \right) = 3\hspace{0.75in}f\left( 1 \right) = 6\]
Sketch a graph of a function that satisfies each of the following conditions. \[\mathop {\lim }\limits_{x \to \, - {3^{\, - }}} f\left( x \right) = 1\hspace{0.75in}\mathop {\lim }\limits_{x \to \, - {3^{\, + }}} f\left( x \right) = 1\hspace{0.75in}f\left( { - 3} \right) = 4\]
Sketch a graph of a function that satisfies each of the following conditions. \[\begin{array}{cll}\mathop {\lim }\limits_{x \to \, - {5^{\, - }}} f\left( x \right) = - 1\hspace{0.5in} & \mathop {\lim }\limits_{x \to \, - {5^{\, + }}} f\left( x \right) = 7\hspace{0.5in} & f\left( { - 5} \right) = 4\\ \mathop {\lim }\limits_{x \to 4} f\left( x \right) = 6\hspace{0.5in} & f\left( 4 \right)\,\,\,{\mbox{does not exist}} & \end{array}\]
Explain in your own words what each of the following equations mean. \[\mathop {\lim }\limits_{x \to {8^{\, - }}} f\left( x \right) = 3\hspace{0.75in}\mathop {\lim }\limits_{x \to {8^{\, + }}} f\left( x \right) = - 1\]
Suppose we know that $\mathop {\lim }\limits_{x \to \, - 7} f\left( x \right) = 18$. If possible, determine the value of $\mathop {\lim }\limits_{x \to \, - 7{\,^ - }} f\left( x \right)$ and the value of $\mathop {\lim }\limits_{x \to \, - 7{\,^ + }} f\left( x \right)$. If it is not possible to determine one or both of these values explain why not.
Suppose we know that $f\left( 6 \right) = - 53$. If possible, determine the value of $\mathop {\lim }\limits_{x \to \,6{\,^ - }} f\left( x \right)$ and the value of $\mathop {\lim }\limits_{x \to \,6{\,^ + }} f\left( x \right)$. If it is not possible to determine one or both of these values explain why not.