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The above procedure shows why taking the pseudoinverse is not a continuous operation: if the original matrix has a singular value 0 (a diagonal entry of the matrix above), then modifying slightly may turn this zero into a tiny positive number, thereby affecting the pseudoinverse dramatically as we now have to take the ...
respectively. If these limits exist at p and are equal there, then this can be referred to as the limit of f(x) at p. [7] If the one-sided limits exist at p, but are unequal, then there is no limit at p (i.e., the limit at p does not exist). If either one-sided limit does not exist at p, then the limit at p also does not exist.
A functor G : C → D is said to lift limits for a diagram F : J → C if whenever (L, φ) is a limit of GF there exists a limit (L′, φ′) of F such that G(L′, φ′) = (L, φ). A functor G lifts limits of shape J if it lifts limits for all diagrams of shape J. One can therefore talk about lifting products, equalizers, pullbacks, etc.
Limits can be difficult to compute. There exist limit expressions whose modulus of convergence is undecidable. In recursion theory, the limit lemma proves that it is possible to encode undecidable problems using limits. [14] There are several theorems or tests that indicate whether the limit exists. These are known as convergence tests.
The function in example 1, a removable discontinuity. Consider the piecewise function = {< = >. The point = is a removable discontinuity.For this kind of discontinuity: The one-sided limit from the negative direction: = and the one-sided limit from the positive direction: + = + at both exist, are finite, and are equal to = = +.
exists and is finite (Titchmarsh 1948, §1.15). The value of this limit, should it exist, is the (C, α) sum of the integral. Analogously to the case of the sum of a series, if α = 0, the result is convergence of the improper integral. In the case α = 1, (C, 1) convergence is equivalent to the existence of the limit
By Darboux's theorem, the derivative of any differentiable function is a Darboux function. In particular, the derivative of the function (/) is a Darboux function even though it is not continuous at one point. An example of a Darboux function that is nowhere continuous is the Conway base 13 function.
Contrary to Fatou's lemma, this value is strictly less than the integral of the limit (0). As discussed in § Extensions and variations of Fatou's lemma below, the problem is that there is no uniform integrable bound on the sequence from below, while 0 is the uniform bound from above.