Real Analysis

The limit of $f(x)$ as $x$ approaches 0 does not exist because the function approaches both positive and negative infinity as $x$ approaches 0 from the positive and negative sides, respectively.

The Cauchy-Schwarz inequality states that for any two vectors $\mathbf{u}$ and $\mathbf{v}$ in an inner product space, the following inequality holds: $\langle \mathbf{u}, \mathbf{v} \rangle^2 \leq \langle \mathbf{u}, \mathbf{u} \rangle \langle \mathbf{v}, \mathbf{v} \rangle$.

A function $f(x)$ is continuous at a point $c$ if the following limit exists and is equal to $f(c)$: $\lim_{x \to c} f(x) = f(c)$.

The function $f(x) = \frac{1}{x}$ is not continuous at $x = 0$ because the limit of the function as $x$ approaches 0 does not exist.

The Intermediate Value Theorem states that if a function $f(x)$ is continuous on a closed interval $[a, b]$, then for any value $y$ between $f(a)$ and $f(b)$, there exists a number $c$ in $[a, b]$ such that $f(c) = y$.

The Mean Value Theorem states that if a function $f(x)$ is continuous on a closed interval $[a, b]$ and differentiable on an open interval $(a, b)$, then there exists a number $c$ in $(a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

A sequence ${a_n}$ is convergent if there exists a real number $L$ such that for any $\varepsilon > 0$, there exists a natural number $N$ such that $|a_n - L| < \varepsilon$ for all $n > N$.

The sequence $a_n = \frac{n}{n+1}$ is convergent because its limit is 1. The sequence $a_n = \frac{(-1)^n}{n}$ is convergent because its limit is 0. The sequences $a_n = \sin(n)$ and $a_n = \cos(n)$ are not convergent because they do not have a limit.

A sequence ${a_n}$ is Cauchy if for any $\varepsilon > 0$, there exists a natural number $N$ such that $|a_m - a_n| < \varepsilon$ for all $m, n > N$.

The sequence $a_n = \frac{n}{n+1}$ is Cauchy because for any $\varepsilon > 0$, we can choose $N = \frac{1}{\varepsilon} + 1$. Then, for all $m, n > N$, we have $|a_m - a_n| = |\frac{m}{m+1} - \frac{n}{n+1}| = \frac{|m(n+1) - n(m+1)|}{(m+1)(n+1)} = \frac{|m-n|}{(m+1)(n+1)} < \frac{1}{N(N+1)} < \varepsilon$.

A metric space $(X, d)$ is complete if every Cauchy sequence in $X$ converges to a point in $X$.

The set of real numbers with the usual metric is complete. The set of rational numbers with the usual metric is not complete. The set of integers with the usual metric is not complete. The set of complex numbers with the usual metric is complete.

A set $S$ in a metric space $(X, d)$ is compact if it is closed and totally bounded. A set $S$ is closed if its complement is open. A set $S$ is totally bounded if for every $\varepsilon > 0$, there exists a finite number of open balls of radius $\varepsilon$ that cover $S$.

The closed interval $[0, 1]$ in the real numbers with the usual metric is compact. The open interval $(0, 1)$ in the real numbers with the usual metric is not compact. The set of rational numbers in the real numbers with the usual metric is not compact. The set of integers in the real numbers with the usual metric is not compact.

A function $f(x)$ is continuous at a point $c$ if the following limit exists and is equal to $f(c)$: $\lim_{x \to c} f(x) = f(c)$.