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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