3целых 7\35 - ( 23\35 - 3\7 ) ) 2целых 5\6 + 7 целых 5\18 − (3 целых 7\9 − 5 18)

Пошаговое объяснение:

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А) Пример  показывает, что первое утверждение не следует из условия. В самом деле, все члены последовательности с четными номерами лежат в любой окрестности точки 1, а предела у последовательности нет, поскольку члены с нечетными номерами не лежат в окрестности (-1;3) точки 1 (по определению предела все члены последовательности начиная с некоторого должны лежать в этой окрестности). 

б) Второе утверждение справедливо, так как если предел равен 5, то все члены последовательности, начиная с некоторого обязаны лежать в окрестности (4;6) точки 5, и они никак не могут оказаться, скажем,  в окрестности (0;2) точки 1.