My book defines uniform continuity as a form of continuity that works for any points $a$ and $x$ in an interval $I$ such that $$|x-a| < \delta$$ implies that $$f(x) – f(a) < \epsilon$$ It then goes on to assert that “If $f$ is continuous over a closed and bounded interval $[a,b]$, it is uniformly […]

So, I’m told I did something wrong on this epsilon-n proof but I really don’t know what. I’m told to prove that $$\lim_{n\to\infty} \frac{5n+1}{n+3} = 5 $$ I begin by setting $\epsilon > 0$ and searching for $k$ such that for every $ n > k$, $$\left|\frac{5n+1}{n+3} – 5\right| < \epsilon$$ $$\left|\frac{5n+1}{n+3} – 5\right| = […]

I couldn’t find an epsilon-delta proof for continuity of $e^x$ so here’s my take: Suppose $|x – x_0| < \delta$ and fix $\epsilon >0$ Consider $|e^x – e^{x_0}| < \epsilon$, then \begin{gather} -\epsilon < e^x – e^{x_0} < \epsilon \\ e^{x_0} – \epsilon < e^x < e^{x_0} + \epsilon \\ ln(e^{x_0} – \epsilon) < x […]

Prove $$\lim_{x\to0}\sqrt{4-x}=2$$ using the precise definition of limits. (Epsilon-Delta) I am not sure how to link $0<\left |x \right |<\delta$ with $\left |\sqrt{4-x}-2\right |<\epsilon$ . EDIT (Trying it out now) I worked till here, then I basically got stuck.

In the $(\epsilon , \delta)$-definition of the limit, $$\lim_{x \to c} f(x) = L,$$ let $f(x) = x^3 + 3x^2 -x + 1$ and let $c = 2$. Find the least upper bound on $\delta$ so that $f(x)$ is bounded within $\epsilon$ of $L$ for all sufficiently small $\epsilon > 0$. I know the definition […]

Most of what I am asking is based off this (fairly popular) article I’ve read here : https://bobobobo.wordpress.com/2008/01/20/how-to-do-epsilon-delta-proofs-1st-year-calculus/, but most lecturers, use this same process to tackle epsilon-delta proofs, so what I am asking should be pretty universal to epsilon delta proofs. Epsilon-Delta Definition of a Limit I’ve just referenced this here, for some added […]

In Spivak’s Calculus Chapter 1 Question 23: Replace the question marks in the following statement ing $\varepsilon, x_0$ and $y_0$ so that the conclusion will be true: If $y_0\neq 0$ and $$|y-y_0|<? \qquad\text{and}\qquad |x-x_0|<?$$ then $y\neq 0$ and $$ \bigg| \frac{x}{y}-\frac{x_0}{y_0}\bigg|<\varepsilon.$$ The answer in its Solution Manual is $$|x-x_0|<\min\bigg(\frac{\varepsilon}{2(1/|y_0|+1)},1 \bigg)$$ and $$|y-y_0|<\min\bigg( \frac{|y_0|}{2},\frac{\varepsilon|y_0|^2}{4(|x_0|+1)} \bigg).$$ The […]

Consider the continuous functions $f,g:\mathbb{R}\rightarrow\mathbb{R}$. Show that $F:\mathbb{R}\rightarrow\mathbb{R}$ with $x\mapsto \max\{f(x),g(x)\}$ is continuous using the $\epsilon – \delta$ definition of continuity. I know there must be four cases. If $f(x)\leq g(x)$ and $f(x_0)\leq g(x_0)$ or if $g(x)\leq f(x)$ and $g(x_0)\leq f(x_0)$ it is easy. However, assuming $f(x_0)\neq g(x_0)$, what if $g(x)\leq f(x)$ and $f(x_0)\leq g(x_0)$ […]

It’s been a while since I’ve studied $\epsilon$-$\delta$ proofs so I’m trying to get a good understanding of how to go about solving this. Working backwards using this as a reference: $\epsilon$-$\delta$ proof that $\lim\limits_{x \to 1} \frac{1}{x} = 1$. Solve: $\lim\limits_{x \to a}$ $\frac{1}{x}$ = $\frac{1}{a}.$ We assume: $\lim_{x\to a}f(x)=\ell \neq 0$. And have: […]

I’m trying to prove that $\lim_{x \to x_0} \frac{1}{ x^2 } = \frac{1}{ {x_0}^2 }$. I know this means that for all $\epsilon > 0$, I must show that there exists a $\delta > 0$ such that $\left | x – x_0 \right | < \delta \Rightarrow \left | \frac{1}{ x^2 } – \frac{1}{ {x_0}^2 […]

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