| Front | [latex]Let $f: A \rightarrow \mathbb{R}$ and $g: B \rightarrow \mathbb{R}$ satisfy $f(A) \subseteq B$ so that the composition $g \circ f$ is defined. Show that if $f$ is differentiable at $c \in A$ and $g$ is differentiable at $f(c) \in B$, then $g \circ f$ is differentiable at $c$ with $(g \circ f)'(c) = g'(f(c)) \cdot f'(c)$.[/latex] |
| Back | [latex]\ \ \ First, notice that since $g$ is differentiable at $f(c)$, it follows that $g'(f(c)) = \lim_{y \rightarrow f(c)} \frac{g(y) - g(f(c))}{y - f(c)}$ for all $y \in B$. Now, let $h(y) = \frac{g(y) - g(f(c))}{y - f(c)}$. Then $g(y) - g(f(c)) = h(y) (y - f(c))$ for all $y \in B$, including $y = f(c)$. If $x \neq c$ for all $x \in A$ where $y = f(x)$ then, $\frac{g(f(x)) - g(f(c))}{x - c} = h(f(x)) \cdot \frac{f(x) - f(c)}{x - c}$. By the Algebraic Limit Theorem, $\lim_{x \rightarrow c} \frac{g(f(x)) - g(f(c))}{x - c} = \lim_{x \rightarrow c} \frac{g(f(x)) - g(f(c))}{f(x) - f(c)} \cdot \lim_{x \rightarrow c} \frac{f(x) - f(c)}{x - c}$. Thus $(g \circ f)'(c) = g'(f(c)) \cdot f'(c)$ and therefore $g \circ f$ is differentiable at $c$.[/latex] |
| Tags | Proof |
| Front | [latex]Let $g: \mathbb{R} \rightarrow \mathbb{R}$, where $$g(x) = \begin{cases}1& \text{if } x \in \mathbb{Q} \\0& \text{if } x \in \mathbb{I}, \text{ where } \mathbb{I} = \mathbb{R} \setminus \mathbb{Q}\end{cases}$$ be the Dirichlet function. Show that $g(x)$ does not have a limit at any real number.[/latex] |
| Back | [latex]\ \ \ Let $c \in \mathbb{R}$ and $L \in \mathbb{R}$. Also, let $\epsilon = \max \{ |L|, |1 - L| \}$, where $\epsilon > 0$. Now, let $\delta > 0$. If $x \in V_\delta (c) \setminus \{ c \}$ and $x \in \mathbb{Q}$, then $f(x) = 1$. If $x \in V_\delta (c) \setminus \{ c \}$ and $x \in \mathbb{I}$, then $f(x) = 0$. The $\epsilon$-neighborhood $V_\epsilon (L)$ of $L$ cannot contain both $0$ and $1$, so $L$ is not a limit at $c$.[/latex] |
| Tags | Proof |
| Front | [latex]Convergence of the p-series[/latex] |
| Back | [latex]The series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if and only if $p > 1$.[/latex] |
| Tags | Theorem |