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## Hot answers tagged epsilon-delta

5

Given $\varepsilon>0$, your formula $\delta:= \min\{1, \frac{\varepsilon}{9}\}$ works! Let's verify it. If $|x-1|<\delta$ with $\delta>0$, then $$|3x^2+1-4|=|3(x-1)(x-1+2)|\leq 3|x-1|(|x-1|+2)<3\delta(\delta+2).$$ Now, by the above definition, $\delta\leq 1$ AND $\delta\leq \frac{\varepsilon}{9}$ which implies $$3\delta(\delta+2)<3\cdot \... 5 There is a thought process here which might feel circular, but the proof itself is not. Keep in mind that I'm allowed to use however silly a thought process I want - the only thing that matters is whether the actual proof produced at the end of the day is valid. If I got my inspiration for what \delta should be by rolling dice, well, that might not be ... 3 Yes, you use the composition theorem$$ f:(x,y)\to xy,\ g:z\to e^z $$with$$ f: \mathbb{R}^2\to \mathbb{R},\ g:\mathbb{R}\to \mathbb{R} $$For the mental exercise, it is very rewarding and I can understand it. Then I advise you to decompose this composition (i.e. general result but with \epsilon,\delta). For example, first for a given \epsilon, find ... 3 You need to bound f(x,y) appropriately from above. So, let \epsilon > 0:$$|f(x, y)| = \frac{x^2}{\sqrt{x^2 + y^2}}\leq \frac{x^2+y^2}{\sqrt{x^2 + y^2}} = \sqrt{x^2+y^2} \stackrel{!}{<}\epsilon$$. So, for \boxed{\delta(\epsilon) = \epsilon} you get for all (x,y) \neq (0,0) with \sqrt{x^2+y^2} < \delta(\epsilon) the desired inequality |... 3 Hint: Use that$$\left|\frac{1}{n}+\frac{\sin(n)}{n}\right|\le \frac{2}{n}$$since we have \left|\sin(n)\right|\le 1 3 |\frac {1+\sin\, n} n| \leq \frac 2 n<\epsilon if n >\frac 2 {\epsilon}. 2 Let \epsilon >0 and assume that the \delta that we are going to choose is less than 1 Note that if |x-4|<\delta and \delta <1 then we have 3<x<5 Thus for x< 4,$$|f(x)-7|=|x^2-9-7|=|x^2-16|=|x+4||x-4|<9|x-4|<9\delta$$On the other hand for x\ge 4,$$ |f(x)-7|=|\frac {2x^2-5x+4}{x-4}-7|=|2x-1-7|=|2x-8|=...

2

The general idea for such $\epsilon$-$\delta$ proofs is that you want to get an upper bound on $|f(x,y) - 0|$, which you can make $< \epsilon$. One possibility is like I hinted at in the comments: Notice that for any $(x,y) \in \Bbb{R}^2$, \begin{align} |x| = \sqrt{x^2} \leq \sqrt{x^2 + y^2} \end{align} Hence, if $(x,y) \neq 0$, then $\dfrac{|x|}{\sqrt{x^... 2 How about polar? Get$\dfrac {r^2\cos^2\theta}r=r\cos^2\theta\le r\to0$. Or you could do it without:$\vert\dfrac {x^2}{\sqrt{x^2+y^2}}\vert\le\vert\dfrac {x^2}{\sqrt {x^2}}\vert=\vert\dfrac {x^2}x\vert=\vert x\vert$, which means we can take$\delta=\epsilon $. 2 Note that we have $$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\lt\left|\frac{xy}{\sqrt{y^2}}\right|=|x|$$ hence the limit exists and equals zero. 2 Your proof is perfect and well explained. I would say it is a matter of personal preference to choose one or the other proof. 2 Or alternatively, by AM-GM we get $${x^2+y^2}\geq 2|xy|$$ so $$\frac{x^2y^2}{x^2+y^2}\le \frac{x^2y^2}{2|xy|}=\frac{1}{2}|xy|$$ and this tends to zero if$x,y$tend to zero. 2 Epsilon - delta proofs can seem circular when you first meet them, but they are certainly not. The point of the proof is this: show that for any number$\epsilon>0$we could choose, there exists a corresponding$\delta>0$so that for every$x$with$|x-a|<\delta$we have$|f(x) - L|<\epsilon$. To do this, we choose any arbitrary$\epsilon$and ... 1 For Dirichlet function, do you understand that$\mathbb{Q}$is dense in$\mathbb{R}$? If so, you will find the function is discontinuous for every point just by$\epsilon-\delta$approach. For the second function, you can just pick any$\epsilon$, then pick$\delta=\epsilon$, for$x\in(-\delta,\delta)$, if$x$is rational,$f(x)=0<\epsilon$, if$x$is ... 1 First, let us look at the$\epsilon-\delta$definition of continuity. If$f$is a real valued function, then it is said to be continuous at a point$x_0 \in \mathbb{R}$if $$\forall \epsilon > 0, \exists \delta > 0 \text{ such that } \forall x \in \mathbb{R} \text{ with } \left| x - x_0 \right| < \delta, \text{ we have } \left| f \left( x \right) - ... 1 Remember what epsilon-delta continuity is: anytime someone gives an epsilon, you have to respond with a delta that works if you want to prove your function is continuous. Sometimes the strategy can feel a little contrived, but imagining this sort of game can be a real help. When dealing with strange functions, think about the "special features" of your ... 1 When someone is working out a proof, there is often a period of time during which they take formulas that look like the desired result and manipulates them in order to find some other formulas that eventually help them write a proof. But nothing they do during that time is part of the proof; it merely helps them figure out a good way to take some step when ... 1 In two variables the epsilon-delta definition for \lim_{\substack{x\to a\\ y\to b}}f(x,y)=L means that for every \epsilon >0 there exists a \delta>0 such that \big|f(x,y)-L\big|<\epsilon whenever 0<\sqrt{(x-a)^2+(y-b)^2}<\delta. In your case, you want to show that \big|f(x,y)-0\big|<\epsilon whenever 0<\sqrt{x^2+y^2}<\... 1 You should be more clear in what you want to show. By definition you have to show that for every \varepsilon > 0 it exists N\in\mathbb{N} such that for every n\geq N it is |x_n-a|<\varepsilon. So let \varepsilon >0 be arbitrary. Then we have to find for that given \varepsilon our N such that the inequality holds. \left|\frac{2n - ... 1 Now, we need$$|3n-7|>\frac{2}{3\epsilon},$$which gives that it's enough to take$$n>\frac{7}{3}+\frac{2}{9\epsilon}.$$1 Polar coordinate gives you$$\lim_{(x,y)\to (0,0)} {xy\over \sqrt{x^2+y^2}}=\lim_{r\to 0} {r^2 \sin \theta \cos \theta\over r} =\lim _{r\to 0} r\sin \theta \cos \theta =0$$1 Well, |x|<2 implies 1/|x|>2, and your conclusion that \left|\frac{6(x-1)}x\right|<|6\delta/2| doesn't hold. Instead, notice that around x=1,x+3>1 so 1/|x+3|<1. Thus,$$\left|\frac{6(x-1)}{x+3}\right|<|6(x-1)|<6\delta<\varepsilon$$giving \delta<\varepsilon/6. 1 x < 2 \Rightarrow \big|\frac{6(x-1)}{x}\big|>\big|\frac{6(x-1)}{2}\big|, not x < 2 \Rightarrow \big|\frac{6(x-1)}{x}\big|<\big|\frac{6(x-1)}{2}\big| 1 If |x-1|<1 then 0<x<2 which implies |x+3|=x+3>3 and$$\left|\frac{8x}{x+3}-2\right| =\frac{6|x-1|}{|x+3|}< \frac{6|x-1|}{3}=2|x-1|.$$Therefore it suffices to take \delta:=\min\{1,\frac{\epsilon}{2}\} 1$$\left|\frac{x^2}{\sqrt{x^2+y^2}}\right|\leq|x|.$$Now, use the$\epsilon-\delta.\$

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