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Natural Log : seems as a discontinous function |
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| Jan22-13, 05:09 AM | #1 |
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Natural Log : seems as a discontinous function
The function f is continuous at some point c of its domain if the limit of f(x) as x approaches c through the domain of f exists and is equal to f(c). In mathematical notation, this is written as
[itex]lim_{x\rightarrow c}[/itex] f(x) = f(c) from the positive and negative sides . For ln(x) (the natural log of x), as x[itex]\rightarrow[/itex]0 , ln(x) approaches -[itex]\infty[/itex] Hence I would stand by the notion that ln(x) is not continuous since at x=0, the function is not defined . Also, the graph does not exist in x<0 domain ; so ln(x) can never approach -[itex]\infty[/itex] from the -ve "x" side . |
| Jan22-13, 05:22 AM | #2 |
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A function is continuous if it continuous at every point in its domain. The domain of natural logarithm (when treated as a real function) is ##(0,\infty)##. Zero is not part of the domain of log.
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| Jan22-13, 05:22 AM | #3 |
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The part you seem to forget is "some point c of its domain". 0 is not in the domain of ln so this is why your observation is not a problem. The domain of ln is the positive numbers.
It is however an interesting observation in its own right and it implies that you cannot possibly define ln(0) in such a way that the extension would be continuous. |
| Jan22-13, 06:42 AM | #4 |
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Natural Log : seems as a discontinous function
extending the idea you proposed, would it be logical to assume that each and every log(x) function has a domain (0,infinity) ?
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| Jan22-13, 06:48 AM | #5 |
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This is a slight lie because in complex analysis we actually extend log and define it for many complex numbers. However we will never be able to include 0 in the domain (without making it discontinuous), and this little side remark is not important to you right now. |
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