For removable discontinuity why limit exists proof
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The limit of a function f at a point c (its existence and its value if it exists) is completely determined by the function values of f near c but not at c.
Also,the limit tries to "predict" f(c) based on the neighboring values of f and it predicts a real number L if the function can be made continuous by setting f(c)=L.
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If the function factors and the bottom term cancels, the discontinuity at the x-value for which the denominator was zero is removable, so the graph has a hole in it.
For example, this function factors as shown:
If the function factors and the bottom term cancels, the discontinuity at the x-value for which the denominator was zero is removable, so the graph has a hole in it.
: After canceling, it leaves you with x – 7. Therefore x + 3 = 0 (or x = –3) is a removable discontinuity — the graph has a hole, like you see in Figure a.
The graph of a removable discontinuity leaves you feeling empty, whereas a graph of a nonremovable
After canceling, it leaves you with x – 7. Therefore x + 3 = 0 (or x = –3) is a removable discontinuity — the graph has a hole, like you see in Figure a.
The graph of a removable discontinuity leaves you feeling empty, whereas a graph of a nonremovable
The graph of a removable discontinuity leaves you feeling empty, whereas a graph of a nonremovable discontinuity leaves you feeling jumpy.
If a term doesn’t cancel, the discontinuity at this x value corresponding to this term for which the denominator is zero is nonremovable, and the graph has a vertical asymptote.
The following function factors as shown:
Because the x + 1 cancels, you have a removable discontinuity at x = –1 (you’d see a hole in the graph there, not an asymptote). But the x – 6 didn’t cancel in the denominator, so you have a nonremovable discontinuity at x = 6. This discontinuity creates a vertical asymptote in the graph at x = 6. Figure b shows the graph of g(x).