Answer to Question #184535 in Real Analysis for Leonard

Question

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

Answer 1

a.

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

b.

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