Exercise 4.2.10

Question

Introductory calculus courses typically refer to the \emph{right-hand limit} of a function as the limit obtained by ``letting $x$ approach $a$ from the right-hand side.”
   \begin{enumerate}[label=(\alph*)] 
       \item Give a proper definition in the style of Definition 4.2.1 for the right-hand and left-hand limit statements:
       $$\lim_{x \to a^+} f(x) = L\text{ and }\lim_{x\to a^-} f(x) = M$$
       \item Prove that $\lim_{x\to a} f(x) = L$ if and only if both the right and left-hand limits equal $L$.
    \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item Let $f : A \rightarrow \mathbf{R}$, and let $c$ be a limit point of the domain $A$. We say that $\lim_{x \to c^+}f(x) = L$ provided that, for all $\epsilon > 0$, there exists a $\delta >0$ such that whenever $0 < x - c < \delta$ (and $x \in A$) it follows that $|f(x) - L| < \epsilon$. We say that $\lim_{x \to c^-}f(x) = L$ provided that, for all $\epsilon > 0$, there exists a $\delta >0$ such that whenever $0 < c - x < \delta$ (and $x \in A$) it follows that $|f(x) - L| < \epsilon$. \item $(\implies)$ If $\lim_{x \to a} f(x) = L$ then for any $\epsilon > 0$, there exists a $\delta > 0$ so that $0 < |x - c| < \delta$ implies $|f(x) - L| < \epsilon$. Since both $0 < x - c < \delta$ and $0 < c - x < \delta$ will satisfy the requirement that $0 < |x - c| < \delta$, then $\lim_{x \to a^+} f(x) = \lim_{x \to a^-} f(x) = L$. $(\impliedby)$ For a given $\epsilon > 0$, there exists $\delta_1, \delta_2 > 0$ so that either $0 < x - c < \delta_1$ or $0 > x - c > -\delta_2$ implies $|f(x) - L| < \epsilon$. If $0 < |x - c| < \delta = \min\{\delta_1, \delta_2\}$ then at least one of the preconditions is always true, so $\lim_{x\to a}f(x) = L$. \end{enumerate}

Exercise 4.2.10

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Exercise 4.2.10

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