Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 6E from Chapter 2.2 from Stewart's Calculus, 8th Edition.
Problem 6E
Chapter:
Problem:
For the function h whose graph is given, state the value of each quantity, if it exists. If it does not exist, explain why.
Step-by-Step Solution
Given information
We are given with a graph of function
We have to find limits
Step 1: (a)
From the graph we can see that the function reaches 4 from left side at x=-3.
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to - {3^ - }} h\left( x \right) = 4\]
Step 2: (b)
From the graph we can see that the function reaches 4 from right side at x=-3.
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to - {3^ + }} h\left( x \right) = 4\]
Step 3: (c)
As the left hand limit and right hand limit are same at x=-3
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to - {3 }} h\left( x \right) = 4\]
Step 4: (d)
The function value at $x=-3$ does not exist, hence
Therefore, the limit is: \[h\left( { - 3} \right) = \,\,\,{\rm{DNE}}\]
Step 5: (e)
From the graph we can see that the function reaches 1 from left side at x=0.
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to {0^ - }} h\left( x \right) = 1\]
Step 6: (f)
From the graph we can see that the function reaches -1 from right side at x=0.
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to {0^ + }} h\left( x \right) = -1\]
Step 7: (g)
As the left hand limit and right hand limit are not same at x=0, so the limit at x=0 does mnot exist
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to 0} h\left( x \right) = {\rm{DNE}}\]
Step 8: (h)
The x coordinate corresponding to $h(x)=1$ is 0,
Therefore, \[{\rm{h}}\left( 0 \right) = 1\]
Step 9: (i)
From the graph we can see that the function reaches 2 from both sides at x=2.
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to 2} h\left( x \right) = {\rm{2}}\]
Step 10: (j)
The function value at x=2 does not existTherefore, the limit is: \[{\rm{h}}\left( 2 \right) = {\rm{DNE}}\]
Step 11: (k)
From the graph we can see that the function reaches 3 from right sides at x=5.
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to {5^ + }} h\left( x \right) = {\rm{3}}\]
Step 11: (l)
From the graph we can see that the function oscillates at x=5. so the limit from left does not exist.
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to {5^ - }} h\left( x \right) = {\rm{DNE}}\]
We are given with a graph of function
We have to find limits
Step 1: (a)
From the graph we can see that the function reaches 4 from left side at x=-3.
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to - {3^ - }} h\left( x \right) = 4\]
Step 2: (b)
From the graph we can see that the function reaches 4 from right side at x=-3.
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to - {3^ + }} h\left( x \right) = 4\]
Step 3: (c)
As the left hand limit and right hand limit are same at x=-3
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to - {3 }} h\left( x \right) = 4\]
Step 4: (d)
The function value at $x=-3$ does not exist, hence
Therefore, the limit is: \[h\left( { - 3} \right) = \,\,\,{\rm{DNE}}\]
Step 5: (e)
From the graph we can see that the function reaches 1 from left side at x=0.
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to {0^ - }} h\left( x \right) = 1\]
Step 6: (f)
From the graph we can see that the function reaches -1 from right side at x=0.
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to {0^ + }} h\left( x \right) = -1\]
Step 7: (g)
As the left hand limit and right hand limit are not same at x=0, so the limit at x=0 does mnot exist
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to 0} h\left( x \right) = {\rm{DNE}}\]
Step 8: (h)
The x coordinate corresponding to $h(x)=1$ is 0,
Therefore, \[{\rm{h}}\left( 0 \right) = 1\]
Step 9: (i)
From the graph we can see that the function reaches 2 from both sides at x=2.
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to 2} h\left( x \right) = {\rm{2}}\]
Step 10: (j)
The function value at x=2 does not existTherefore, the limit is: \[{\rm{h}}\left( 2 \right) = {\rm{DNE}}\]
Step 11: (k)
From the graph we can see that the function reaches 3 from right sides at x=5.
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to {5^ + }} h\left( x \right) = {\rm{3}}\]
Step 11: (l)
From the graph we can see that the function oscillates at x=5. so the limit from left does not exist.
Therefore, the limit is: \[\mathop {\lim }\limits_{x \to {5^ - }} h\left( x \right) = {\rm{DNE}}\]