Answer to Question #184535 in Real Analysis for Leonard

Question #184535
  1. Use the definition of the limit of a sequence to establish
  • lim ( 3n2+1 / 6n2+2 ) = 1/2
  • lim ( (n+2)1/2 - (n)1/2 ) = 0
1
Expert's answer
2021-04-28T11:00:53-0400

a.

"Let \\space \\epsilon >0 \\space be \\space given.\\newline\n\\begin{vmatrix}\n \\frac{3n^2+1}{6n^2+2}-\\frac{1}{2}\n\\end{vmatrix}=\\begin{vmatrix}\n \\frac{3n^2+1}{2(3n^2+1)}-\\frac{1}{2}\n\\end{vmatrix}\n=0\\newline\nTake\\space \\delta=\\epsilon.\\newline\nSo, when \\begin{vmatrix}\n n-0\n\\end{vmatrix}<\\epsilon=\\delta\\newline\n \\implies, \\begin{vmatrix}\n \\frac{3n^2+1}{6n^2+2}-\\frac{1}{2}\n\\end{vmatrix}<\\epsilon\\newline\n\\implies lim_{n \\to 0}\\frac{3n^2+1}{6n^2+2}=\\frac{1}{2}"


b.

"Let \\space \\epsilon>0 \\space be\\space given.\\newline\\begin{vmatrix}\n \\sqrt{n+1}-\\sqrt{n}-0\n\\end{vmatrix}\\hspace{2cm}Rationalise \\space the\\space denominator\\newline\n=\\begin{vmatrix}\n \\frac{1}{\\sqrt{n+1}+\\sqrt{n}}\n\\end{vmatrix}\n=\\frac{1}{\\sqrt{n+1}+\\sqrt{n}}<\\frac{1}{\\sqrt{n}+\\sqrt{n}}=\\frac{1}{2\\sqrt{n}}<\\frac{1}{\\sqrt{n}}<\\epsilon\\newline\nIf\\space \\sqrt{n}>\\frac{1}{\\epsilon}\\newline\n\\hspace{0.8cm} {n}>\\frac{1}{\\epsilon^2}\\newline\nIf\\space m \\space is\\space a\\space \\space positive \\space integer>\\frac{1}{\\epsilon^2},\\space then\\newline\n\\hspace{1cm} \\begin{vmatrix}\n \\sqrt{n+1}-\\sqrt{n}-0\n\\end{vmatrix}<\\epsilon\\hspace{1.5cm}\\forall n\\geq m \\newline\n\\implies lim_{n\u21920}\u200b\\sqrt{n+1}-\\sqrt{n}\u200b=0"

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