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The function = { < = > has no limit at x 0 = 1 (the left-hand limit does not exist due to the oscillatory nature of the sine function, and the right-hand limit does not exist due to the asymptotic behaviour of the reciprocal function, see picture), but has a limit at every other x-coordinate.
A limit of a sequence of points () in a topological space is a special case of a limit of a function: the domain is in the space {+}, with the induced topology of the affinely extended real number system, the range is , and the function argument tends to +, which in this space is a limit point of .
On the other hand, if X is the domain of a function f(x) and if the limit as n approaches infinity of f(x n) is L for every arbitrary sequence of points {x n} in X − x 0 which converges to x 0, then the limit of the function f(x) as x approaches x 0 is equal to L. [10] One such sequence would be {x 0 + 1/n}.
In this case, a single limit does not exist because the one-sided limits, and + exist and are finite, but are not equal: since, +, the limit does not exist. Then, x 0 {\displaystyle x_{0}} is called a jump discontinuity , step discontinuity , or discontinuity of the first kind .
in an essential discontinuity, oscillation measures the failure of a limit to exist. This definition is useful in descriptive set theory to study the set of discontinuities and continuous points – the continuous points are the intersection of the sets where the oscillation is less than ε (hence a G δ set ) – and gives a very quick proof ...
Here, a is not required to lie in the interior of I. Indeed, if a is an endpoint of I, then the above limits are left- or right-hand limits. A similar statement holds for infinite intervals: for example, if I = (0, ∞), then the conclusion holds, taking the limits as x → ∞. This theorem is also valid for sequences.
The pointwise limit of a sequence of continuous functions may be a discontinuous function, but only if the convergence is not uniform. For example, f ( x ) = lim n → ∞ cos ( π x ) 2 n {\displaystyle f(x)=\lim _{n\to \infty }\cos(\pi x)^{2n}} takes the value 1 {\displaystyle 1} when x {\displaystyle x} is an integer and 0 {\displaystyle ...
If () for all x in an interval that contains c, except possibly c itself, and the limit of () and () both exist at c, then [5] () If lim x → c f ( x ) = lim x → c h ( x ) = L {\displaystyle \lim _{x\to c}f(x)=\lim _{x\to c}h(x)=L} and f ( x ) ≤ g ( x ) ≤ h ( x ) {\displaystyle f(x)\leq g(x)\leq h(x)} for all x in an open interval that ...