Answer to Question #184136 in Real Analysis for Maria

Question #184136

i) Show by using an (ε − N ) argument that

lim 3n2 −2n+1 = 3

n→∞ 2n2 − 4 2


ii) Use an (ε − δ) argument to show that f : R → R be the function defined by


􏰂x2−5x−5 if x≥−1 f(x)= x2+x+1 if x<−1


is continuous at x = −1.


1
Expert's answer
2021-05-04T14:36:40-0400

"\\displaystyle\n(1)\\\\\\lim_{n \\to \\infty} \\frac{3n^2 - 2n + 1}{2n^2} = \\frac{3}{2}\\\\\n\n\\textsf{We must show that}\\,\\, \\forall \\epsilon > 0,\\,\\,\\exists \\,\\,N(\\epsilon) \\in \\mathbb{N},\n\\\\ \\textsf{such that}\\,\\, n > N(\\epsilon) \\implies |a_n - a_0| < \\epsilon\\\\\n\n\\begin{aligned}\n|a_n - a_0| &= \\left|\\frac{3n^2 - 2n + 1}{2n^2} - \\frac{3}{2}\\right| = \\left|\\frac{3n^2 - 2n + 1 - 3n^2}{2n^2}\\right|\n\\\\&= \\left|\\frac{-2n + 1}{2n^2}\\right|\n\\end{aligned}\\\\\n\n-2n + 1 < -2n + 2 < 2n - 2 < 2n\\\\\n\n\\begin{aligned}\n|a_n - a_0| = \\left|\\frac{-2n + 1}{2n^2}\\right| < \\left|\\frac{2n}{2n^2}\\right| = \\left|\\frac{1}{n}\\right|\n\\end{aligned}\\\\\n\n\\implies -\\epsilon < \\frac{1}{n} < \\epsilon\\\\\n\\implies n > \\frac{1}{\\epsilon}\\\\\n\n\\therefore \\exists \\textsf{a}\\,\\, N(\\epsilon) \\in N \\,\\, \\textsf{such that}\\\\\nn > N(\\epsilon) \\implies |a_n - a_0| < \\epsilon \\,\\,\\textsf{which verifies the limit.}\\\\\n\n(2)\\\\\nf(x) = x^2\u22125x\u22125\\,\\, \\textsf{if}\\,\\, x \\geq\u22121,\\,\\, f(x) = x^2+x+1\\,\\, \\textsf{if}\\,\\, x<\u22121\\\\\n\nf(-1) = (-1)^2 - 5(-1) - 5 = 1\\\\\n\n\\textsf{We must show that if}\\,\\, |x - (-1)| < \\delta \\implies |x + 1| < \\delta,\\\\ |f(x) - f(-1)|< \\epsilon\n\n|f(x) - f(-1)| = |x^2 - 5x - 5 - 1| < \\epsilon\\\\\n\n\\begin{aligned}\n|x^2 - 5x - 6| &< |x^2 + 2x + 1|\n\\\\\\implies|x^2 + 2x + 1| &< \\epsilon\n\\\\\\implies |(x + 1)^2| &< \\epsilon \\implies |x + 1| < \\sqrt{\\epsilon}\n\\end{aligned}\\\\\n\n\\implies \\delta = \\sqrt{\\epsilon}\\\\\n\n\\textsf{This implies that}\\,\\, |f(x) - f(-1)|< \\epsilon\\\\\n\nf(-1) = (-1)^2 + (-1) +1 = 1\\\\\n\n\\textsf{We must show that if}\\,\\, |x - (-1)| < \\delta \\implies |x + 1| < \\delta, \\\\\n|f(x) - f(-1)|< \\epsilon\n\n|f(x) - f(-1)| = |x^2 + x - 2| < \\epsilon\\\\\n\n\\begin{aligned}\n|x^2 + x - 2| &< |x^2 + 2x + 1|\n\\\\\\implies|x^2 + 2x + 1| &< \\epsilon\n\\\\\\implies |(x + 1)^2| &< \\epsilon \\implies |x + 1| < \\sqrt{\\epsilon}\n\\end{aligned}\\\\\n\n\\implies \\delta = \\sqrt{\\epsilon}\\\\\n\n\\textsf{This implies that}\\,\\, |f(x) - f(-1)|< \\epsilon \\\\\n\n\\textsf{Hence,}\\,\\, f(x)\\,\\, \\textsf{is continuous at}\\,\\, x = 1."


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