endsem 2018 - (Q 1-A-a Let{an be a sequence of real numbers If{an is convergent then prove that its limit is unique[3 marks Answer Let if possible the

endsem 2018 - (Q 1-A-a Let{an be a sequence of real numbers...

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(Q. 1-A-a.) Let { a n } be a sequence of real numbers. If { a n } is convergent then prove that its limit is unique. [3 marks] Answer: Let, if possible, the sequence { a n } converge to two different real numbers, say L 1 and L 2 , and let 3 δ = | L 1 - L 2 | . [1 mark] Then for δ we have N 1 , N 2 N such that | L i - a n | < δ n N i . [1 mark] By choosing N to be the maximum of N 1 and N 2 , we get a contradiction, thus L 1 = L 2 . [1 mark] (Q. 1-B-a.) Let { b n } be a sequence of real numbers. If { b n } is convergent then prove that its limit is unique. [3 marks] (Q. 1-C-a.) Let { c n } be a sequence of real numbers. If { c n } is convergent then prove that its limit is unique. [3 marks] (Q. 1-D-a.) Let { d n } be a sequence of real numbers. If { d n } is convergent then prove that its limit is unique. [3 marks] (Q. 1- · -b.) State the sandwich theorem for sequences. [3 marks] Answer: Let { a n } , { b n } and { c n } be sequences of real numbers such that [1 mark] a n b n c n n. Assume that { a n } and { c n } are convergent with lim n →∞ a n = lim n →∞ c n = α , say. [1 mark] Then (the sequence { b n } is also convergent and) lim n →∞ b n = α . [1 mark] (Note: The convergence of the middle sequence, { b n } , is a part of the conclusion here. If you assume it in the beginning then your statement is incorrect.)
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(Q. 2-A.) Consider the function f ( x ) = ( 7 - (16) 1 x 7+(16) 1 x if x 6 = 0 7 if x = 0 . Prove that there is a point c such that 2 < c < 4 and f ( c ) = 1 2 . [6 marks] Answer: The function is continuous in the interval (2 , 4) , because the exponential function is continuous on the whole of R and 1 /x is continuous for x 6 = 0 . [2 marks] Consider g ( x ) := f ( x ) - 1 2 , and note that g is also continuous on (2 , 4) . Observe that g (2) = f (2) - 1 2 = 7 - 4 7 + 4 - 1 2 < 0 and g (4) = f (4) - 1 2 = 7 - 2 7 + 2 - 1 2 > 0 . [ 1 mark each ] Therefore by the intermediate value theorem, there is some c (2 , 4) such that g ( c ) = 0 , or f ( c ) = 1 2 . [2 marks] (Q. 2-B.) Consider the function f ( x ) = ( 5 - (16) 1 x 5+(16) 1 x if x 6 = 0 5 if x = 0 . Prove that there is a point c such that 2 < c < 4 and f ( c ) = 1 4 . [6 marks] (Q. 2-C.) Consider the function f ( x ) = ( 9 - (16) 1 x 9+(16) 1 x if x 6 = 0 9 if x = 0 . Prove that there is a point c such that 2 < c < 4 and f ( c ) = 1 2 . [6 marks] (Q. 2-D.) Consider the function f ( x ) = ( 11 - (16) 1 x 11+(16) 1 x if x 6 = 0 11 if x = 0 . Prove that there is a point c such that 2 < c < 4 and f ( c ) = 1 2 . [6 marks]
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(Q. 2-A.) Consider the function f ( x ) = ( 7 - (16) 1 x 7+(16) 1 x if x 6 = 0 7 if x = 0 . Prove that there is a point c such that 2 < c < 4 and f ( c ) = 1 2 . [6 marks] Answer: Note that f ( c ) = 1 2 if and only if 7 3 c = 16 . [2 marks] Note that ( 7 3 ) 2 < 16 < ( 7 3 ) 4 as 2 < 7 3 < 4 . [2 marks] The function x 7→ ( 7 3 ) x is continuous and by intermediate value theorem a required c (2 , 4) exists. [2 marks] (Q. 2-B.) Consider the function f ( x ) = ( 5 - (16) 1 x 5+(16) 1 x if x 6 = 0 5 if x = 0 . Prove that there is a point c such that 2 < c < 4 and f ( c ) = 1 4 . [6 marks] (Q. 2-C.) Consider the function f ( x ) = ( 9 - (16) 1 x 9+(16) 1 x if x 6 = 0 9 if x = 0 . Prove that there is a point c such that 2 < c < 4 and f ( c ) = 1 2 . [6 marks] (Q. 2-D.) Consider the function f ( x ) = ( 11 - (16) 1 x 11+(16) 1 x if x 6 = 0 11 if x = 0 .
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