A function is continuous at a point n, if we have;
1) [tex]\lim\limits_{x \to n} f(x)[/tex] is defined
2) [tex]\lim\limits_{x \to n} f(x)[/tex] exist, or has a value
3) [tex]\lim\limits_{x \to n} f(x) = \mathbf{f(n)}[/tex]
The true statement is statement (a)
(a) True
Statements b, c, d, e, f, g, h, i, and j are false
(b) False, (c) False, (d) False, (e) False, (f) False, (g) False, (h) False, (i) False
The reason the above selection are correct is as follows:
The given parameters are;
The domains and ranges of the function are:
To the left: Domain; 1 ≤ x < 0, Range; 1 ≤ f(x) < 0
To the right: Domain; 0 < x < 1, Range; 0 < f(x) < 1
The function is discontinuous at x = 0, and x = 1
The statements are analyzed individually as follows:
(a) [tex]\lim\limits_{x \to -1^+} f(x) = 1[/tex] True
The reason the above statement is true is that a domain of the function is 1 ≤ x < 0, -1 is the left boundary of the function
Therefore as x tends to -1 from the right, which is -1⁺, the function is defined, which is f(-1) = 1
(b) [tex]\lim\limits_{x \to 0^-} f(x) = 0[/tex] False
An open hole on a graph, indicates that the function is discontinuous at the value of x = 0
Therefore, as x tends to 0, f(x) is not defined
(c) [tex]\lim\limits_{x \to 0^-} f(x) = 1[/tex] False
The function is discontinuous at the value of x = 0
(d) [tex]\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x)[/tex] False
There is a discontinuity at x = 0
(e) [tex]\lim\limits_{x \to 0^-} f(x)[/tex] exist False
At x = 0, the function is discontinuous
(f) [tex]\lim\limits_{x \to 0} f(x) = 0[/tex] False
At x = 0, the f(x) is discontinuous
(g) [tex]\lim\limits_{x \to 0} f(x) = 1[/tex] False
From the graph at x = 0 f(x) is discontinuous
(h) [tex]\lim\limits_{x \to 1} f(x) = 1[/tex] False
From the graph, [tex]\lim\limits_{x \to 1} f(x)[/tex] Does not exist
(i) [tex]\lim\limits_{x \to 1} f(x) = 0[/tex] False
[tex]\lim\limits_{x \to 1} f(x)[/tex] Does not exist
(j) [tex]\lim\limits_{x \to 2^-} f(x) = 2[/tex] False
The function is not defined at x = 2⁻
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