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### 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 ...
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### 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_{...
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### $\lim_{x\to 1} \sqrt{x^2 + 8} = 3$ -- 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 $0<\delta<1$. $1-\delta < x < 1+\delta$ $1+\delta^2-2\delta < x^2<1+2\delta +\delta^2$...
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### $\lim_{x\to 1} \sqrt{x^2 + 8} = 3$ -- 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 ...
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### 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 ...
• 50.2k
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### 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 $x_n=\frac{1}{n}$...
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Accepted

• 1,481
1 vote

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 $\frac{1}{\delta}$-$\frac{1}{\... • 663 1 vote ### Epsilon delta proof :$\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| &... 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 ... • 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)=...
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