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$\epsilon$-$\delta$ proof, finding $\delta$

Could we have have used the inequality: for all positive $x$ less than $1$, $$|\log(x)| \leq \dfrac{1}{\sqrt{x}}.$$ I guessed it from the fact that $$\lim\limits_{x\to 0^{+}} \sqrt{x}\log(x)=0.$$ Then,...
Yathiraj Sharma's user avatar
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$\epsilon$-$\delta$ proof, finding $\delta$

I think you can easily, changing $x$ by $\lfloor x\rfloor$, reduce the asked (write, if not) case to a fraction $\frac{\log _{2}n}{n}, n\in \mathbb{N}$ and let me show some technique for its ...
zkutch's user avatar
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1 vote

Epsilon proof verification

The calculation is all fine, but it falls short at the end. Setting $\varepsilon$ to be a function of $K$ is a misstep; you don't get to determine what form $\varepsilon$ takes, as it is an arbitrary ...
Theo Bendit's user avatar
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3 votes
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For $a>1$ and a fixed $N$ does $\sum\limits_{n=1}^\infty \prod\limits_{k=0}^{n-1} \left(1- \frac{a}{a+N +k} \right)$ converge?

Multiply your series by $$\Gamma(N)\Gamma(a)/ \Gamma(a+N)$$ and your series becomes $$\sum_{n=1}^{\infty}\frac{\Gamma(a)\Gamma(N)\prod_{k=0}^{n-1} (N+k)}{\Gamma(a+N)\prod_{k=0}^{n-1} (a+N+k)}=\sum_{n=...
Kroki's user avatar
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Proving Discontinuity using Epsilon Delta for Function

We want to prove the discontinuity at $0$, i.e. $$\exists\epsilon>0,\forall\delta>0,\exists x\in[-\delta,\delta]\setminus\{0\},\left|\frac1x-c\right|\ge\epsilon.$$ For this, any $\epsilon>0$ ...
1 vote

Show discontinuity with $\epsilon-\delta$

An easier way would to be argue by sequential definition of continuity. If we can find a sequence ${x_n}\to 0$ but $f(x_n)\not\to f(0)=c$ then we're done. But if you pick the sequence $x_n=\frac{1}{n}$...
J.Dmaths's user avatar
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2 votes

Show that $\lim_{x\to 4} \sqrt{x} = 2$ using $\epsilon$-$\delta$ definition of limit.

HINT Another possible approach. Let $|x - 4| < \delta_{\varepsilon}$. Then one concludes that: \begin{align*} |\sqrt{x} - 2| = \frac{|x - 4|}{\sqrt{x} + 2} < \frac{|x - 4|}{2} < \frac{\delta_{...
Átila Correia's user avatar
1 vote
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Show that $\lim_{x\to 4} \sqrt{x} = 2$ using $\epsilon$-$\delta$ definition of limit.

Scratch Work The goal is to determine a value of $\delta > 0$ such that $|x-4| < \delta$ implies that $|\sqrt{x} - 2| < \varepsilon$. Note $$ \bigl| x-4 \bigr| = \bigl| (\sqrt{x}-2)(\sqrt{x}+...
Xander Henderson's user avatar
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Show that $\lim_{x\to 4} \sqrt{x} = 2$ using $\epsilon$-$\delta$ definition of limit.

As noted in the comments, both your methods are wrong due to a wrong choice of $\delta$. One thing I find helpful when trying to proof these kind of limits is to choose multiple $\delta$ for different ...
watertrainer's user avatar
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Proving that Jacobian of Composition is equal to Composition of Jacobians using epsilon-delta

Here is a literal translation of my proof of the chain rule in my "Fundamentos de calculo" (in Spanish), which I can post here since I own the rights. In the vector case, $\varepsilon$-$\...
William M.'s user avatar
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2 votes
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Why Is This Jump Discontinuity Not Differentiable?

$\lim_{h\to 0}\frac{2(x+h)+1-(2x+1)}{h}=2$ only proves that for $x>0$, $f'(x)=2$. Likewise, $\lim_{h\to 0}\frac{2(x+h)-2x}{h}=2$ only proves that for $x<0$, $f'(x)=2$. They prove nothing about $...
Anne Bauval's user avatar
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2 votes

Proof that f is not continuous

The definition of limit is that for any $\epsilon > 0$, there exists $\delta > 0$ such that whenever $0 < |x-a| < \delta$, then $|f(x) - L| < \epsilon$. The idea is that if there ...
heropup's user avatar
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Epsilon delta proofs on discontinuous functions

If $\lim\limits_{x \rightarrow 0}f(x) = 0$ , then given $\epsilon>0$, there must be $\delta>0$ such that $0<|x-0|<\delta, x \neq 0 \implies |f(x)-0|<\epsilon$ Now, for $x \neq 0$, $f(x)=...
Ruan Carlos's user avatar
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Epsilon delta proof with x going towards infinity.

You need to show that for $\epsilon >0 $ there is an $M$, such that $x > M$ implies $|h(x) - (-2)| <\epsilon.$ (Comment by J. Douma) Note: Let $x>0:$ $|h(x) - (-2)|=\dfrac{2}{e^{2x}+1}<\...
Peter Szilas's user avatar
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1 vote

Epsilon delta proof with x going towards infinity.

You need to be cautious: What you want to show is that $$|h(x) - (-2)| < ε$$ for $x$ large; this is not what you are given! To do this, fix $ε>0$. We have $$|h(x)-(-2)| = \frac{\frac{2}{e^x}}{e^...
Tobius's user avatar
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Epsilon delta proof with x going towards infinity.

As you said $\dfrac{\dfrac{2}{e^x}}{e^x+\dfrac{1}{e^x}}<\epsilon$ That's equivalent to saying $\frac{2}{e^{2x}+1}<\epsilon$ so $e^{2x}$+1>$\frac{2}{\epsilon}$, so $e^{2x}>\frac{2}{\epsilon}...
J.Dmaths's user avatar
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Is this definition of limits better than epsilon-delta definition of limits?

Well, for a start , how do we define one sided limits with this definition? Furthermore, what if we take f(x) =$\frac{1}{x}$ and L=0 at c=0. It is automatically true that $\frac{1}{\delta}$-$\frac{1}{\...
J.Dmaths's user avatar
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Having trouble with proving $\lim_{z\to0}\frac{(Re(z))^2}{|z|} = 0$ using the epsilon definition

As an alternative we can also use that by $z=x+iy$ and $x\neq 0$ $$\frac{(\Re(z))^2}{|z|}=\frac{x^2}{\sqrt{x^2+y^2}}\le \frac{x^2}{|x|}=|x|\to 0$$
user's user avatar
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Why is the condition $0<|x-a|<\delta$ necessary in the $\varepsilon$-$\delta$ definition of a limit?

"capturing the idea of closeness inherent in the concept of a limit" : This is exactly the point, which can be formalized in modern infinitesimal analysis as follows: whenever $\alpha$ is a ...
Mikhail Katz's user avatar
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1 vote

Epsilon delta proof : $\lim_{x \rightarrow 0}e^\frac{-1}{x^2}=0$

We want to show that for all $\epsilon > 0$, there exists $\delta > 0$ s.t. $$|x| < \delta \implies \big|e^\frac{-1}{x^2}\big| < \epsilon$$ Suppose $|x| < \delta$, $$\begin{align} |x| &...
Ahmed Elrefaey's user avatar
6 votes

Why is the condition $0<|x-a|<\delta$ necessary in the $\varepsilon$-$\delta$ definition of a limit?

EDIT: In response to your edit, yes, your last explanation is right--the concept of "closeness" is the whole point of the limit definition. If $x$ is "close" to $a$, then $f(x)$ is ...
Pavan C.'s user avatar
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1 vote
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Basic analysis question to estimate the growth of a function

Let's say you got the result for $t \in (-\delta, \delta) \cup (-\infty, A) \cup (B, \infty)$ with for example $A \leq -\delta$, $\delta \leq B$ to make the next expression easy to write. Now, the ...
Bruno B's user avatar
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8 votes
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If $\lim\limits_{x\to0}f(x)=0$ and $\lim\limits_{x \to 0}\frac{f(2x)-f(x)}{x}=0$ how to rigorously prove that $\lim\limits_{x \to 0}\frac{f(x)}{x}=0$?

Since $\lim_{x \rightarrow 0} f(x) = 0$ you can write $f(x)$ as a telescoping sum $$f(x) = \sum_{n=0}^{\infty} (f(2^{-n} x) - f(2^{-n-1}x))$$ $$ = \sum_{n=0}^{\infty} (2^{-n - 1}x) {f(2^{-n} x) - f(2^{...
Zarrax's user avatar
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1 vote
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Is there a different way of proving continuity of $g(x) = \max \{f(t) \lvert 0 \leq t \leq x\}$

For $0\leq x<y$ let $M(x,y)=\sup_{x\leq t\leq y} f(t)$, so that $g(x)=M(0,x)$. Then $g(y)=\max(g(x),M(x,y))$. Observe by continuity of $f$ that $\lim_{y\to x}M(x,y)=f(x)$. Thus $$ \lim_{y\to x}g(y)=...
pre-kidney's user avatar
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