To be point continuous the function has to exist at that point and when evaluated at that point it must equal the two one sided limits at that point. For 2, that is
lim[x↑4] √x = lim[x↓4] √x = √4
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NB: x↑4 is the left handed limit, "x increases to 4" and [x↓4] is the right handed limit "x decreases to 4"
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This is obviously true because all of them equal 2 and therefore the function is continuous at c=4.
7.
Always start with finding whether the function exists at the point. In this case it obviously doesn't because f(2) = 0/0 which is not allowed. The limits, however, do exist and this ratio can be simplified, so this function is discontinuous, but only at c=2. This is called a removable discontinuity, or a hole.
11.
This is obviously not continuous at c=1 because the function doesn't exist at the point, f(1) = 1/0. Also the limit doesn't exist because the one sided limits do not equal each other. This is called a non-removable discontinuity and is typified by a vertical asymptote.
lim[x↑4] √x = lim[x↓4] √x = √4
§§
NB: x↑4 is the left handed limit, "x increases to 4" and [x↓4] is the right handed limit "x decreases to 4"
§§
This is obviously true because all of them equal 2 and therefore the function is continuous at c=4.
7.
Always start with finding whether the function exists at the point. In this case it obviously doesn't because f(2) = 0/0 which is not allowed. The limits, however, do exist and this ratio can be simplified, so this function is discontinuous, but only at c=2. This is called a removable discontinuity, or a hole.
11.
This is obviously not continuous at c=1 because the function doesn't exist at the point, f(1) = 1/0. Also the limit doesn't exist because the one sided limits do not equal each other. This is called a non-removable discontinuity and is typified by a vertical asymptote.