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evagelos
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how can we prove that lun x is continuous over (0, [tex]\infty[/tex] )?
Provided that we define : lun x =y <=> [tex]e^y =x[/tex]?
Provided that we define : lun x =y <=> [tex]e^y =x[/tex]?
mathman said:lnx - lnu = ln(x/u). For any fixed x > 0, u->x => x/u -> 1 and ln1=0.
You can dress this proof up.
Continuity is a concept in mathematics that describes the behavior of a function at a specific point. A function is considered continuous if it is defined at that point and there are no sudden jumps or breaks in its graph.
To prove continuity of a function, you must show that the function is defined at a specific point, the limit of the function exists at that point, and the limit is equal to the value of the function at that point.
Continuity and differentiability are related concepts, but they are not the same. Continuity refers to the behavior of a function at a specific point, while differentiability refers to the existence of the derivative of a function at a point.
Yes, a function can be continuous at a point but not differentiable. This can occur if the function has a sharp corner or cusp at that point, which would make the derivative undefined.
To find the continuity of a function on a closed interval, you must check the continuity of the function at each endpoint of the interval and at every point in between. If the function is continuous at all of these points, then it is considered continuous on the closed interval.