Exercise 4.2.9

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[Infinite Limits] The statement $\lim _{x \rightarrow 0} 1 / x^{2}=\infty$ certainly makes intuitive sense. To construct a rigorous definition in the challenge response style of Definition $4.2 .1$ for an infinite limit statement of this form, we replace the (arbitrarily small) $\epsilon>0$ challenge with an (arbitrarily large) $M>0$ challenge: \emph{Definition:} $\lim _{x \rightarrow c} f(x)=\infty$ means that for all $M>0$ we can find a $\delta>0$ such that whenever $0<|x-c|<\delta$, it follows that $f(x)>M$. \begin{enumerate}[label=(\alph*)] \item Show $\lim _{x \rightarrow 0} 1 / x^{2}=\infty$ in the sense described in the previous definition. \item Now, construct a definition for the statement $\lim _{x \rightarrow \infty} f(x)=L$. Show $\lim _{x \rightarrow \infty} 1 / x=0 .$ \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item For a given $M > 0$, if $0 < |x - 0| = |x| < 1/\sqrt{M} = \delta$ then $1/|x|^2 = 1/x^2 < M$ as desired. \item $\lim_{x \to \infty}f(x) = L$ means that for all $\epsilon > 0$ we can find a $N$ such that when $x > N$ it follows that $\left|f(x) - L\right| < \epsilon$. For a given $\epsilon > 0$, choosing $N = 1/\epsilon$ leaves us with $x > N \implies 1 / N = \epsilon > 1/x$ hence $\lim_{x \to \infty} 1/x = 0$. \end{enumerate}

Exercise 4.2.9

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Exercise 4.2.9

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