Question

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.

Answer 1

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