Question 1 :
Sketch the graph of a function f that satisfies the given values :
f(0) is undefined
lim x -> 0 f(x) = 4
f(2) = 6
lim x -> 2 f(x) = 3
Solution :
From the given question,
The value of x approaches from left and right, the limit will approach the value 4.
When x approaches 2 from left and right, the limit will approaches to 3. The picture given above will illustrate the condition.
When x = 2, f(x) that is the value of y will be 6.
Hence the picture given above is the required graph of the statements given.
Question 2 :
Sketch the graph of a function f that satisfies the given values :
f(-2) = 0
f(2) = 0
lim x -> -2 f(x) = 0
lim x -> 2 f(x) does not exists
Solution :
From the given question,
When x = -2, the value of y will be 0.
When x = 2, the value of y will be 0.
When x tends to 2, the function does not exist. To show this, we have to show the graph with different values of y.
Question 3 :
Write a brief description of the meaning of the notation lim x -> 8 f(x) = 25
Solution :
When x approaches from left side and right side, the value of limit will approaches 25.
lim x -> 8- f(x) = 25 |
lim x -> 8+ f(x) = 25 |
Question 4 :
If f(2) = 4, can you conclude anything about the limit of f(x) as x approaches 2?
Solution :
The given statement represents, when x = 2, the value of y will be 4.
Case 1 :
When x approaches from left side and right side, we will get same values approximately, or
Case 2 :
When x approaches from left side and right side, we will get different values.
Hence, we cannot conclude anything about the limit of f(x) as x approaches 2?
Question 5 :
If the limit of f(x) as x approaches 2 is 4, can you conclude anything about f(2)? Explain reasoning.
Solution :
Given :
lim x -> 2 f(x) = 4
From this, we may understand that
lim x -> 2- f(x) = 4 |
lim x -> 2+ f(x) = 4 |
when x approaches 2 from left side and right side, the limit will approaches to 4.
Hence we cannot conclude anything about f(2).
Question 6 :
Evaluate :
lim x->3 (x2 -9)/(x - 3) if it exists by finding f(3+) and f(3-)
Solution :
= im x->3 (x2 -9)/(x - 3)
= im x->3 (x + 3)(x - 3)/(x - 3)
= im x->3 (x + 3)
im x->3+ f(x) = 3 + 3 = 6 |
im x->3- f(x) = 3 + 3 = 6 |
Question 7 :
Verify the existence of lim x -> 1 f(x)
Solution :
If the limit x -> 1 exists, then
lim x-> 1- f(x) = lim x-> 1+ f(x)
f(x) = (x - 1)/(x - 1) f(x) = 1 lim x-> 1+ f(x) = 1 |
f(x) = -(x - 1)/(x - 1) f(x) = -1 lim x-> 1- f(x) = -1 |
Since lim x-> 1- f(x) ≠ lim x-> 1+ f(x), the limit does not exists.
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