1. –/2 points scalcet7 2.2.002.nvaMy Notes Explain what it means to say that lim
ID: 2889842 • Letter: 1
Question
1.–/2 pointsscalcet7 2.2.002.nvaMy Notes
Explain what it means to say that
lim x 9f(x) = 6 and lim x 9+f(x) = 3.
As x approaches 9, f(x) approaches 3, but f(9) = 6.
As x approaches 9 from the right, f(x) approaches 6. As x approaches 9 from the left, f(x) approaches 3.
As x approaches 9 from the left, f(x) approaches 6. As x approaches 9 from the right, f(x) approaches 3.
As x approaches 9, f(x) approaches 6, but f(9) = 3.
In this situation is it possible that
lim x 9 f(x)
exists? Explain.
Yes, f(x) could have a hole at (9, 6) and be defined such that f(9) = 3.
Yes, f(x) could have a hole at (9, 3) and be defined such that f(9) = 6.
Yes, if f(x) has a vertical asymptote at x = 9, it can be defined such that lim x9f(x) = 6, lim x9+f(x) = 3, and lim x9 f(x) exists.
No, lim x9 f(x) cannot exist if lim x9f(x) lim x9+f(x).
2.–/2 pointsscalcet7 2.2.ae.007.nvaMy Notes
EXAMPLE 7 The graph of a function g is shown in the figure. Use it to state the values (if they exist) of the following:
(a) lim x 2g(x)
(b) lim x 2+g(x)
(c) lim x 2 g(x)
(d) lim x 5g(x)
(e) lim x 5+g(x)
(f) lim x 5 g(x).
SOLUTION From the graph we see that the values of g(x) approach as x approaches 2 from the left, but they approach as x approaches 2 from the right. Therefore
(a) lim x 2g(x) = and (b) lim x 2+g(x) = .
(c) Since the left and right limits are different, we conclude that the limit as x approaches 2 of g(x) does not exist.
The graph also shows that
(d) lim x 5g(x) = and (e) lim x 5+g(x) = .
(f) This time, the left and right limits are the same and so, by this theorem, we have
lim x 5 g(x) =
Despite this fact, notice that g(5) 4.
3.–/2 pointsscalcet7 2.2.004.nvaMy Notes
Use the given graph of f to state the value of each quantity, if it exists. (If an answer does not exist, enter DNE.)
(a)
lim x 2f(x)
(b)
lim x 2+f(x)
(c)
lim x 2 f(x)
(d)
f(2)
(e)
lim x 4 f(x)
(f)
f(4)
4.–/2 pointsscalcet7 2.2.006.nvaMy Notes
For the function h whose graph is given, state the value of each quantity, if it exists. (If an answer does not exist, enter DNE.)
(a)
lim x 3h(x)
(b)
lim x 3+h(x)
(c)
lim x 3 h(x)
(d)
h(3)
(e)
lim x 0h(x)
(f)
lim x 0+h(x)
(g)
lim x 0 h(x)
(h)
h(0)
(i)
lim x 2 h(x)
(j)
h(2)
(k)
lim x 5+h(x)
(l)
lim x 5h(x)
5.–/2 pointsscalcet7 2.2.007.nvaMy Notes
For the function g whose graph is given, state the value of each quantity, if it exists. (If an answer does not exist, enter DNE.)
(a)
lim t 0g(t)
(b)
lim t 0+g(t)
(c)
lim t 0 g(t)
(d)
lim t 2g(t)
(e)
lim t 2+g(t)
(f)
lim t 2 g(t)
(g)
g(2)
(h)
lim t 4 g(t)
6.–/2 pointsscalcet7 2.2.008.nvaMy Notes
For the function R whose graph is shown, state the following. (If an answer does not exist, enter DNE.)
(a)
lim x 2 R(x)
(b)
lim x 5 R(x)
(c)
lim x 3R(x)
(d)
lim x 3+R(x)
(e) The equations of the vertical asymptotes.
7.–/2 pointsscalcet7 2.2.010.mi.nvaMy Notes
A patient receives a 150-mg injection of a drug every 4 hours. The graph shows the amount f(t) of the drug in the bloodstream after t hours.
Find
lim t 12f(t)
and
lim t 12+f(t).
lim t 12f(t) = mg
lim t 12+f(t) = mg
8.–/2 pointsscalcet7 2.2.024.mi.nvaMy Notes
Use a table of values to estimate the value of the limit. If you have a graphing device, use it to confirm your result graphically.lim x0
9.–/2 pointsscalcet7 2.2.030.mi.nvaMy Notes
Determine the infinite limit.lim x10
10.–/2 pointsscalcet7 2.2.038.nvaMy Notes
Find the vertical asymptotes of the function.y =
Confirm your answer by graphing the function.
11.–/2 pointsscalcet7 2.2.041.nvaMy Notes
(a) Estimate the value of the limit
lim x0 (1 + x)1/x
to five decimal places.
(b) Illustrate part (a) by graphing the function
y = (1 + x)1/x.
12.–/2 pointsscalcet7 2.2.046.nvaMy Notes
In the theory of relativity, the mass of a particle with velocity v ism =
,where m0 is the mass of the particle at rest and c is the speed of light. What happens as
v c?
m
m
m m0
m 0
Question Part Points Submissions Used xExplanation / Answer
1. Please note that limit of a function will exist only if the left hand limit and the right hand limit of a function are equal.
We are given that the left hand limit for f(x) that is as x approaches 9 from left is = 6
and the right hand limit for f(x) that is as x approaches 9 from the right is = 3
Hence the answer is
No, lim x9 f(x) cannot exist if lim x9f(x) lim x9+f(x).
2. Please note the holes in the grapg imply that the function is not defined at that particular va;ue of x. Like
In our graph at x = 2 the function does not exits.
However as x approaches 2 from the left the function approaches 4 and when x approaches 2 from the right the function approaches 3.
Hence the limit does not exist as x approaches 2
=>
(a) lim x 2g(x) = 4 and (b) lim x 2+g(x) = 3 .
(c) Since the left and right limits are different, we conclude that the limit as x approaches 2 of g(x) does not exist.
Next the solid dot in the graph represent that the value of the function when x = 5 is = 3
And as x approaches 5 from the left the function approaches 4 and when x approaches 5 from the right the value of the function approaches 4.
Hence the limit exists as x aapraches 5 and is = 4
=>
(d) lim x 5g(x) = 4 and (e) lim x 5+g(x) = 4 .
(f) This time, the left and right limits are the same and so, by this theorem, we have
lim x 5 g(x) = 4
Despite this fact, notice that g(5) 4.
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