8 votes
Accepted

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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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
  • 1,242
4 votes

Proving a function isn't bounded

The formulas that you've written actually prove something stronger, namely that $$ \lim_{x \to 0}\; \biggl\lvert \frac1x \biggr\rvert = \infty. $$ To establish that $f(x) = \frac1x$ is unbounded, we ...
Sammy Black's user avatar
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3 votes
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If $f(0)=f(1)=0$ and for every $x\in (0,1)$ exists $0<\delta \inf\{x,1− x\}$ with $f(x)=\frac{f(x−\delta)+f(x+\delta)}2$. How to show that $f(x)=0$?

Suppose $f$ is not identically zero. WLOG we can assume it is positive for some $x$. Then its maximum $M$, which is achieved, is positive. Let $f(x_0)=M$. Let $x_1= \sup\{x: f(x)=M\}$. By continuity, $...
GReyes'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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2 votes
Accepted

Proving a function isn't bounded

If $f: \mathbb{R}\setminus \{0\} \to \mathbb{R}$ was bounded there would exist a constant $M >0$ such that $$ |f(x)| < M, \quad \forall x\ne 0. $$ However, this generates a contradiction since $...
PierreCarre'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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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

$\lim_{x\to 1} \sqrt{x^2 + 8} = 3$ -- Proof Via Epsilon-Delta Definition of a Limit

Prove $\lim_{x\to 1} \sqrt{x^2+8}=3$ We need to find a delta given an epsilon, so we try to find $0<\delta<1$. $1-\delta < x < 1+\delta$ $1+\delta^2-2\delta < x^2<1+2\delta +\delta^2$...
TurlocTheRed's user avatar
  • 5,465
1 vote
Accepted

$\lim_{x\to 1} \sqrt{x^2 + 8} = 3$ -- Proof Via Epsilon-Delta Definition of a Limit

Given $\epsilon>0$, instead of providing some $\delta$ out of the blue and check it "works", like textbooks too often do, we shall look for the "best" $\delta$ (i.e. the largest ...
Anne Bauval'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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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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1 vote
Accepted

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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1 vote

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
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
  • 1,481
1 vote

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
  • 663
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
1 vote
Accepted

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
  • 4,727
1 vote
Accepted

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
  • 30.2k

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