Answer to Question #191547 in Real Analysis for Amara

Question

1. Show by using an (ε − N ) argument that lim n->inf 3n^2 −2n+1/2n^2 − 4 =3/2
2. Use an (ε − δ) argument to show that f : R → R be the function defined by

f(x)= {x^2−5x−5 if x≥−1 x^2+x+1 if x<−1}

is continuous at x = −1

Answer 1

1.)

$x_n={3n^2-2n+1 \over 2n^2-4 }$

$|x_n-\frac{3}{2} |=|\frac{3n^2-2n+1}{2n^2-4}-\frac{3}{2}|$

$=|\frac{6n^2-4n+2-(6n^2-12)}{2(2n^2-4)}|$

$|x_n-\frac{3}{2} |=\frac{|4n-7|}{|4n^2-8|}=\frac{2n-7}{2n^2-4}$ for $n\ge 4$

now

$|x_n-\frac{3}{2} |=\frac{2n-7}{2n^2-4} <\frac{2n}{2n^2-4}=\frac{n}{n^2-2}$

but $n^2-2>\frac{n^2}{2} \ for \ n \ge4$

thus

$|x_n-\frac{3}{2} |<\frac{2n}{n^2}= \frac{2}{n}<\epsilon \ \ \ \ \ if \ n> \frac{2}{\epsilon}$

thus

$|x_n-\frac{3}{2} |=\epsilon \ \ \ \ \ \ \ \ \ for \ all \ n \ge [\frac{2}{\epsilon},4]$

2.)

$f(x) = \begin{cases} x^2-5x-5 \ \ \ \ \ & x \ge-1 \\ x^2+x+1 & x\le-1 \end{cases}$

$f(-1)=1$

let choose $\delta$ such that $0<\delta <1$

thus $|x-1|<\delta \ \ \ \ \ \ \implies |x+1|<1$

thus $|x-6|=|x+1-7|<|x+1|+7<8$

and $|x|=|x+1-1|<|x+1|+1<2<8$

now for $0<\delta <1$ , we have $|x-6|<8 \ , |x|<\delta$

thus $|f(x)-1|<|x+1|.\delta < \epsilon \ \ if \ |x+1|<\frac{\epsilon}{\delta}$

hence choose $\delta$ = min {$\frac{\epsilon}{\delta},$ 1 } , we get

$|f(x)-1|<\epsilon$ for all $|x+1|< \delta$

and $\delta = min$ {$\frac{\epsilon}{\delta},$ 1}