How to Sketch a Graph of a Function With Limits :
Here we are going to see how to sketch a graph of a function with limits.
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 (x^{2} -9)/(x - 3) if it exists by finding f(3^{+}) and f(3^{-})
Solution :
= im _{x->3} (x^{2} -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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