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Real Analysis
Is V=(1,2) a neighbourhood of the point 1.6 ? what epsilon neighbourhood would work ? Five value of epsilon
Real Analysis
Show that the space L^p is a normed linear space
Real Analysis
Prove that if x is not a negative integer , summation n=1 to infinity of 1/(x+n)(x+n+1) = 1/1+x .
Real Analysis
(1) for the function f , defined by f(x) = 4x^3 - 4x^2 -7x -2 , there exist a point c €]-1/2 ,2[ satisfying f'(c) = 0
(2) for all even integral values of n, lim (x+1)^-n as x tends to infinity exists.
(2) for all even integral values of n, lim (x+1)^-n as x tends to infinity exists.
Real Analysis
(1) check whether the function f given by f(x) = 1/(2x-4)^2 for all x belonging to ]-2,2[ is continuous in the interval ]-2,2[. Is it bounded? Justify your answer.
(2) find the values of p and q so that lim { p sinx + x(1-cosx)}/5x^3 as x tends to infinity is 1/6
(2) find the values of p and q so that lim { p sinx + x(1-cosx)}/5x^3 as x tends to infinity is 1/6
Real Analysis
Let a function f : R -> R be defined by f(x) = {2, if x belongs to Q , 4 , if x doesn't belongs to Q} check whether f is continuous on B
Real Analysis
Use a proof of contradiction to establish the following:
If a positive whole number n can be expresses as n1 n2, where n1 is greater equals to 2 and n2 is greater equals to 2, then at least one element sets n1 and n2 is less than n^1/2
If a positive whole number n can be expresses as n1 n2, where n1 is greater equals to 2 and n2 is greater equals to 2, then at least one element sets n1 and n2 is less than n^1/2
Real Analysis
Use a proof bt of the set of n1 and n2 is less than n^1/2y contradiction to establish the following:
If a positive whole number n can be expresses as n1 n2, where n1 is greater equals to 2 and n2 is greater equals to 2, then at least one element sets n1 and n2 is less than n^1/2
If a positive whole number n can be expresses as n1 n2, where n1 is greater equals to 2 and n2 is greater equals to 2, then at least one element sets n1 and n2 is less than n^1/2
Real Analysis
How can I prove that the sequence of irrational numbers is unbounded
Real Analysis
Let G denote the family of all open subsets of the real numbers and F the fam-
ily of all closed subsets of the real numbers. Then
Every set G in G is the union of a sequence of disjoint open intervals (called the compo-
nents of G).
ily of all closed subsets of the real numbers. Then
Every set G in G is the union of a sequence of disjoint open intervals (called the compo-
nents of G).
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