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\date{Due Tuesday, 4/2}
\title{Math 104 Homework 9 (Vaintrob)}
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\section{Reading Exercise 1}
Repeat example 3 (chapter 20) to (rigorously) find the derivative of $\sqrt{x}$ at $\pi,$ i.e.\ $\lim_{x\to \pi} \frac{\sqrt{x} - \sqrt{\pi}}{x-\pi}.$
\section{Reading Exercise 2} The limit of a function $f$ at a real number $a$ (on the boundary of its domain) exists and equals $L$ if and only if $f$ can be extended to a continuous function $\tilde{f}$ on $\text{Dom}(f)\cup \{a\}$ with $\tilde{f}(a) = L$ (see Theorem 19.5). Using this, quickly re-prove Corollary 20.8 (without using Theorem 20.6) using the $\epsilon$-$\delta$ notion of continuity.
\section{Reading Exercise 3} (a) Suppose $f(x)$ is a continuous function on the interval $[a, b]$ and $g(x)$ is a continuous function on the interval $[b,c],$ such that $f(b) = g(b)$ (equal to some value $L$, say). Show that the function defined by $$\tilde{f}(x) : = \begin{cases}f(x)& x\in [a, b]\\ g(x)& x\in [b,c]\end{cases}$$ is continuous on the interval $[a, c].$

(b) Now, look at the condition on $f(x)$ after the ``if and only if'' statement in Theorem 20.10 (very top of p. 161). The statement $\lim_{x\to a^-}f(x) = L$ is equivalent to saying that the function $f(x)$ to the left of $a$ extends to a continuous function on $x\in J\mid x\le a$, with value $L$ at $a$. The statement $\lim_{x\to a^+}f(x) = L$ is equivalent to a similar statement for values to the right of $a$ (you are allowed to use these facts). Deduce that the two conditions together are equivalent to requiring that $f(x)$ extends to a continuous function on all of $J$. (This line of argument can be continued to give an alternative proof of theorem 20.10.)

\section{19.2}
\section{19.8}
\section{20.11}
\section{20.17}


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