Just a thought came in my mind , i understand the original def of epsliom delta version of limit . Why cant the limit definition be based on&quot; for every delta there exits epsilon thing such that...&quot;? Why the mathematicians give def for &quot;every epsilon there exists delta such that..&quot;, was there some kind of contradiction in that def for some  functions where it was not valid? Hence the epsilon to delta was selected ?
123698
So the definition of Limit I see is $$\lim_{x \to a}f(x) = L$$ means: for all $\epsilon &gt;0$, there exists a $\delta &gt;0$ such that $$0&lt;|x - a| &lt; \delta \Rightarrow |f(x) - L| &lt; \epsilon $$ I was wondering if modifying the definition to: Limit exists when for all $\delta &gt;0,$ there exists a $\epsilon &gt; 0$ such that $0&lt;|f(x) - L| &lt; \epsilon $ $\Rightarrow |x - a| &lt; \delta$ would cause any problem. Because to me it seems like this definition should also work except it does seem a lot harder when it comes to the actual proving part. Can this flipped version of the limit definition also work? Why did mathematicians define in this order?
48461
How to prove the following limit?   $$\lim_{n \to \infty} (1+1/n)^n = e$$    I can only observe that the limit should be a very large number!   Thanks.
61967
Give an epsilon-delta proof that the limit for $x\to 0$ of $x^3\sin(x)$ exists.  If I take a really big number for $x$ it doesn't seem to have a limit...
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Prove the following using epsilon-delta definition:  Let $a \in \mathbb{R}$ and $f,g$ be defined on $\mathbb{R}$.  If  $$\lim_{x\to a} f(x) = 647 \tag 1$$  and $$\lim_{x\to a} g(x) = \infty \tag 2$$ then   $$\lim_{x\to a} f(x) + g(x) = \infty.$$  $$(1) \implies\forall \epsilon &gt; 0, \exists \delta &gt; 0, s.t. |x-a|&lt;\delta \implies |f(x)-647| &lt; \epsilon$$  $$(2) \implies \forall M &gt; 0, \exists \delta_2 &gt; 0, s.t. |x-a|&lt;\delta_2 \implies |g(x)| &gt; M$$  Not sure about this... I've set up the definitions. I believe we will have to take a minimum between the deltas for first two limits then use it in the third.  Anyone have any clues?
37404
In my assignment I have to prove that the following function doesn't have a limit:  $$f(x)=\frac{x}{x-\lfloor{\sin x}\rfloor}$$ when $${x \to 0}$$  In other words, I have to prove that for every $\delta$ such that $0&lt;|x-{x_0}|&lt;\delta,$ $|f(x)-L|&gt;\epsilon$.  In fact, I don't even know how to start. Do I choose my $\delta$? I know that the sin function has certain values that's supposed to help me. How do I use it?  I have to prove this by definition, so I can't use limit arithmetic for example  Thanks.
86742
Prove that $\displaystyle \lim_{x \to a} x^2 = a^2$   Let $\varepsilon &gt; 0$, and let $\delta = \min(\frac{\varepsilon}{2|a|+1}, 1)$. Suppose $x \in\mathbb{R} - \left\{a\right\} $ and $|x-a| &lt; \delta.$ Then $|x-a| &lt; 1$ hence $ -1 &lt; x-a &lt;1 $ hence $ a-1 &lt; x &lt; a+1$ therefore $2a-1&lt; x+a &lt;2a+1.$ Thus $ |x+a| &lt; 2|a|+1.$ So $|x^2-a^2| = |x-a||x+a|$ $ \displaystyle &lt; (2|a|+1)\delta =(2|a|+1)\frac{\varepsilon}{(2|a|+1)} = \varepsilon. $  Is this correct? Is there another choice of delta we could have chosen?
55885
I want to use $\epsilon-\delta$ formulation to prove that $\lim_{x\to 1 } 2x+3 \neq 6$.  I know that I need to show that there exists an $\epsilon&gt;0$ such that for every $\delta&gt;0$ there exists $x\in \mathbb{R}$ such that $0&lt;\left| x-1 \right| &lt;\delta$ but $|2x-3|\geq \varepsilon$.  Will someone please guide me through the process?  Thanks a lot in advance
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It is easy to prove the limit exists, all we have to show is there exists a relationship between $\delta$ and $\epsilon$. But how are we supposed to prove limit doesn't exists? The problem is when we are proving for a limit we already know what the limit is and with that, algebra is all that's needed.   Please show through an example (you may show that $\lim_{x\rightarrow0} \frac{1}{x}$ doesn't exist)  If possible please use the explanation scheme that is used by this answer
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Using only the delta definition of a limit, how can we prove that the sequence $\{a_n\}$, where $a_n = \sin n$, as $n$ tends to infinity does not have a limit?  Thanks!
50635
Question    Suppose we are dealing with real valued functions $f(x)$ of one real variable $x$. We say that the limit of $f(x)$ at point $a$ exists and equals to $L$ if and only if  $$\exists L:\left( {\forall \varepsilon  &gt; 0,\exists \delta  &gt; 0:\left( {\forall x,0 &lt; \left| {x - a} \right| &lt; \delta  \implies \left| {f(x) - L} \right| &lt; \varepsilon } \right)} \right)$$  and show this by the symbolism  $$\mathop {\lim f(x)}\limits_{x \to a}  = L$$  Now, what do we say when we want to state that the limit of function $f(x)$ does not exist at $a$? In fact, what is the negation of the above statement? I am interested to obtain the negation with a step by step approach using tautologies in logic. To be specific, I want to start from  $$\neg \left[ {\exists L:\left( {\forall \varepsilon  &gt; 0,\exists \delta  &gt; 0:\left( {\forall x,0 &lt; \left| {x - a} \right| &lt; \delta  \implies \left| {f(x) - L} \right| &lt; \varepsilon } \right)} \right)} \right]$$  and then go through it to get the final form of negation (See the example below).    My Thought  I just wrote down the two following negations without going through a step by step approach.  $$\forall L,\exists \varepsilon  &gt; 0:\left( {\forall \delta  &gt; 0,\exists x:0 &lt; \left| {x - a} \right| &lt; \delta  \implies \left| {f(x) - L} \right| \ge \varepsilon } \right)$$  or  $$\nexists L:\left( {\forall \varepsilon  &gt; 0,\exists \delta  &gt; 0:\left( {\forall x,0 &lt; \left| {x - a} \right| &lt; \delta  \implies \left| {f(x) - L} \right| &lt; \varepsilon } \right)} \right)$$  I want to know weather these are true or not.       Example    I will give an example of what I mean by a step by step approach. Consider the following statement  $${P}\implies R$$  I want to take a step by step approach to obtain the negation of the above statement. Here is what they usually do in logic  \begin{align} \, \neg \left( {P \implies R} \right) &amp;\iff \neg \left( {\neg P \vee R} \right) &amp; \text{Conditional  Disjunction} \\ \qquad \qquad \quad &amp;\iff \neg \neg P \wedge \neg R &amp; \text{Demorgan's  Law} \\ \qquad \qquad \quad &amp;\iff P \wedge \neg R &amp; \text{Double  Negation} \end{align}
32562
Is $f(x)=x+\frac{x}{x+1}$ uniformly continuous on $(0,\infty)$   Going from the epsilon delta definition we get:   $$\forall x,y&gt;1,\text{WLOG}:x&gt;y \ ,\ \forall\epsilon&gt;0,\exists\delta&gt;0 \ s.t. \ |x-y|&lt;\delta\rightarrow |x+\frac{x}{x+1}-y-\frac{y}{y+1}|&lt;\epsilon$$  But I'm not really sure on how to continue from here.  Alternatively: if $\lim\limits_{x\to \infty}f(x)$ exists on $[0,\infty)$ then the function is uniformly continuous, well, it's easy to see here that $\lim\limits_{x\to \infty}f(x)=\infty$ but is that enough ?
26153
Is it trivial to say  $$\mathop {\lim }\limits_{n \to \infty } {(1 + {k \over n})^n} = e^{k},$$  considering the fact that we know $$\mathop {\lim }\limits_{n \to \infty } {(1 + {1 \over n})^n} = e?$$
67170
For example, can we say: $\infty=\lim\limits_{n\rightarrow\infty} n &lt; \aleph_0$?  These are two different types of structures. The limit being like the length, extension, or just generic magnitude and the other being cardinality of a set. Can we compare magnitude to cardinality?  Intuitively, we can reach $\aleph_0$ by counting the natural numbers on the number line and in the process will be approaching $\infty$. Which leads me to believe $\infty\leq\aleph_0$. But I can't see why it should be a strict inequality. I feel like they should be of equal magnitude.  I saw on a recent comment that $2^\infty=\infty$, but are those infinities really the same? It seems not to me. Of course we (usually) have that  $2^{\aleph_0}=\aleph_1$ where ${\aleph_0}$ and $\aleph_1$ are clearly two very distinct infinities, countable vs uncountable at least. Maybe one might argue that as far as the concept of magnitude is concerned, all infinities are "equal".
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I know that it is defined such that there is no more than one y-value output for any given x-value input, but I"m wondering WHY it is defined that way? Why can't we apply everything we know about functions to equations that have more than one y-value output for any given x-value input?   In short, what is the benefit or reason to define a function this way?  At the same time, a little historical context would also be appreciated (i.e., when the concept originated and why).  Thanks,  Moshe
104636
I'd like to ask a question which I have been reflecting on for some time now. What is the limit of: $f(x) = \sin(x)$ as $x$ tends to infinity?  As we know, the function has a definite value for each multiple of a value included between $0$ and $2\pi$, but, how can we know which value it will have at infinity?
95036
I know that $\infty-\infty$ is indeterminate, the how about $\lim_{x\to \infty}(x-x)$ and $\lim_{x\to \infty}x-\lim_{x\to \infty}x$?   Thank you.
39255
I know $R^\omega$ is the set of functions from $\omega$ to $R$. I would think $R^\infty$ as the limit of $R^n$, but isn't that $R^\omega$? The seem to be used differently, but I can't tell exactly how.