How would you do an Epsilon-Delta proof for this theorem?  IF $\lim \limits_{x \to a^+}$  $f(x)$ = $\lim \limits_{x \to a^-}$ $f(x) = L$   THEN $\lim \limits_{x \to a}$  $f(x)$ = L
21688
My teacher proposed this question as a challenge proof to do on our own and I can't seem to get it. Was Wondering if anyone could give me a hint or help me on the process to completing it.  Let $a \in \mathbb{R}$. Let $f$ be a function defined, at least, on an interval  centred at $a$, except possibly at $a$. Let $L \in \mathbb{R}$.     If $\lim\limits_{ x→a-} f(x) = \lim\limits_{ x→a+} f(x) = L$   then $\lim\limits_{ x→a} f(x) $ = $L$.    We are supposed to prove this using the "delta-epsilon" proof style. Thanks in advance!!
84750
Use epsilon-delta definition of limit  To show :  $\displaystyle \lim_{x \to 0}  x\sin \left(\dfrac{1}{x}\right) = 0$   How to explain this in detail with step by step explanations?
24630
I need to prove the  continuity of $f(x)=\log x$ using a $\epsilon-\delta$ proof  These is what I have so far but am not sure how to continue  $|\log x-\log a| &lt; \epsilon$   $\log a- \epsilon &lt; \log x &lt; \log a+ \epsilon$  $\frac{a}{e^\epsilon} &lt; x &lt; {a}e^\epsilon$  Any help is appreciated
18212
Given a sequence $\{x_n\}_{n=1}^\infty$ of positive numbers, satisfying $\lim_{n \to \infty} x_n^{x_n} = 4$, prove $\lim_{n\to\infty}x_n=2$. My attampt: from $\lim_{n \to \infty} x_n^{x_n} = 4$ we know $\exists n_0\in\mathbb{N}:\forall n\ \geq n_0: |x_n^{x_n} -4| &lt; \epsilon$ so $|x_n-2| = \frac{|x_n^2-4|}{|x_n+2|} \leq |x_n^2-4| \leq |x_n^{x_n} -4| &lt;\epsilon$ Which completes the proof if correct. Is it?
25859
I'm really struggling on this question.  I've been thinking about it for awhile now but this is one of those where I don't really have much intuition on what to do.  Problem 3. Prove that if the limit $\lim_{x\to +\infty}f(x) =: L$ exists (finite or infinity) and the improper integral. $$\int_a^{+\infty}f(x) dx$$ is convergent, then $L = 0$.  Here is some facts that I know that I think I should perhaps use. Facts/Knowledge I know: The above integral can be rewritten as:  $$\int_a^{\infty}f(x)dx = \lim_{A\to \infty}\int_a^{A}f(x)dx$$  We know that limit as x goest to infinity of $f(x)=L$.  I need to somehow show that limit L is equal to 0. If I take the integral I get $\lim_{A \to \infty} F(A)-f(a)$ and I know this does not equal infinity as the integral converges.  I'm not sure if this is on the right track or what to do next. I was thinking maybe perhaps instead I could use the Cauchy Criterion to come up with a proof.  Note: I'm not used to doing proofs with Improper integrals.  Theorem 1 (Cauchy Criterion). The improper integral (1) converges if and only if for every $\epsilon &gt; 0$ there is an $M\geqslant a$ so that for all $A, B \geqslant M$ we have $$\Bigg|\int_A^B f(x) dx\Bigg| &lt; \epsilon$$
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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
89629
I'm starting  and finally decided to learn how to write epsilon-delta proofs.  I have been working on chapter 5, number 3(ii). The problem, in essence, asks to prove that     $$\lim\limits_{x \to 1} \frac{1}{x} = 1.$$    Here's how I started my proof,     $$\left| f(x)-l \right|=\left| \frac{1}{x} - 1 \right| =\left| \frac{1}{x} \right| \left| x - 1\right| &lt; \epsilon \implies \left| x-1 \right| &lt; \epsilon |x|$$   I haven't made any further progress past this point. Is it possible to salvage this proof? Should I try an alternate approach?
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I am having a hard time grasping the definition of a limit. I initially learned this loose definition of a limit:  $\lim\limits_{x \to a}f(x)=L$ iff $f(x)$ approaches L when $x$ approaches $a$ from both directions.    Then i learned the epsilon delta definition, which states the following:  $\lim\limits_{x \to a}f(x)=L$ iff for any $\epsilon &gt; 0$ there exists some $\delta&gt;0$ where if $0&lt;|x-a|&lt;\delta$ then $|f(x)-L]&lt;\epsilon$  The way i think about this definition(please correct me if I'm wrong) is that for any given range $\epsilon$ around L, we can find some range $\delta$ around x such that the function evaluated at those points will be within $\epsilon$ of L. However i don't understand how that statement is the same thing as saying:  $\lim\limits_{x \to a}f(x)=L$ iff $f(x)$ approaches L when $x$ approaches $a$ from both directions.
119987
Let $f:(a,b)\to \mathbb{R}$ be a strictly increasing function. Does the limit $\lim_{x\to a^+}f(x)$ necessarily exist and is a real number or $-\infty$?  If so, is it true that $\ell=\lim_{x\to a}f(x)\le f(x) \ \ \forall x\in (a,b)$? Please provide proofs.
27614
Let $f:X \rightarrow Y$ be a function. Prove that if $f$ is continuous, then for every convergent sequence $(x_n)$ lim$_{n\rightarrow \infty}\,f(x_n)=f(\text{lim}_{n \rightarrow \infty} \, x_n)$  My attempt: Assume $f$ is continuous. Then $\forall \epsilon &gt;0, \exists \delta &gt; 0$ such that $\forall p \in X$ and $q \in X: d(p,q) &lt; \delta \implies d(f(q),f(q)) &lt; \epsilon.$  And also assume that $(x_n)$ converges. Then $\exists L$ such that $\forall \epsilon &gt;0, \exists N$ such that $n&gt;N \implies d(x_n, L) &lt; \epsilon.$  I need to use these definitions to show lim$_{n\rightarrow \infty}\,f(x_n)=f(\text{lim}_{n \rightarrow \infty} \, x_n)$ . Any hints?  Edit: lim$_{n \rightarrow \infty}(x_n) \in X$
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Assume that $$\lim_{n\to\infty}(a_{n+2}-a_{n})=A$$ show that  $\displaystyle \lim_{n\to\infty}\dfrac{a_{n}}{n}$and $\displaystyle\lim_{n\to\infty}\dfrac{a_{n+1}-a_{n}}{n}$ exist and find these limits.  maybe this problem have some methods, Thank you.  since  $$\lim_{n\to\infty}(a_{n+2}-a_{n})=A$$ so there exists $N$, and for $\forall \varepsilon&gt;0$, such $$|a_{n+2}-a_{n}-A|\le\varepsilon$$
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I want to prove the definition of limit of a function from definition (1) to definition (2).  (1) A function $f(x)$ is said to have $f^*$ as the limit as $x$ tends $x^*$ iff for every sequence $\{x_n\}$ with limit $x^*$, the limit $\lim_{n-&gt;\infty}f(x_n)$ exists and is equal to $f^*$. (2) A function is said to have $f^*$ as the limit as $x$ tends $x^*$ iff for each $\epsilon &gt;0$, there exists a $\delta_\epsilon &gt;0$ such that $|f(x)-f^*| &lt;\epsilon \forall |x^*-x|&lt;\delta_\epsilon$  I am confused about the steps. Suppose I am given an $\epsilon &gt; 0$ and $x^*$. I have to show the existence of a $\delta_\epsilon &gt;0$ such that for all $x$ satisfying  $|x^*-x|&lt;\delta_\epsilon$ I can construct a sequence $x_n$ which converges to $x^*$ and $x_n=x$ for some $n \geq N$. Then I can say that $f(x_n)-f^*=f(x)-f^* &lt; \epsilon$. I think I may have made some mistake about the steps.  My main confusion is the following : Suppose I choose $\delta_\epsilon=\epsilon$. Then if a sequence $\{x_n\}$ converges to $x^*$ then $x_n=x$ (where $x$ satisfies $|x-x^*|&lt;\epsilon$) for some $n\geq N$. Now, $\lim_{n-&gt;\infty}f(x_n)$ exists implies there exists an $M$ such that $\forall n\geq M$, $f(x_n)-f^* &lt;\epsilon$. Now, I need to show that for some $n \geq M$ indeed $x_n=x$. How do I show that ? This $M$ and $N$ are different.
87774
How do I prove that:   $\lim \limits_{n\to \infty}2^{1/n}=1$  Thank you very much.
102214
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?
74431
I've been stuck on this question for a long time now and was wondering if anyone could show me how it's done. So far I have done the following: Since $\lim_{x\to \infty}f'(x)=M$ then $\forall \epsilon ,\exists A &gt;0$  s.t. if $x&gt;A$ then $|f'(x)-M|&lt;\epsilon$. So we have that $M-\epsilon&lt;f'(x)&lt;M+\epsilon$. By MVT on $[x,x+1]$ for $x&gt;A$, $\exists c\in (x,x+1)$ s.t. ${f(x+1)-f(x)\over 1}=f'(c)$. Therefore, $M-\epsilon&lt;f(x+1)-f(x)&lt;M+\epsilon$. However, I do not know what to do after this.
28033
Proof that $\lim{a_n}=L$ when $n$ goes to infinity, then $\{a_n\}$ its a Cauchy Sequence  I start with this hypothesis  $\displaystyle\lim_{n \to\infty}{a_n}=L \Leftrightarrow{\forall{\epsilon}&gt;0}$  $\exists{N}\in{\mathbb{N}}$ such that  $\forall{n}&gt;N \left |{a_n-L}\right |&lt; \epsilon$  The same its true fot any $m&gt;N,\left |{a_m-L}\right |=\left |{L-a_m}\right |&lt; \epsilon$  I will sum the inequalities but I havent any idea how do it... Help me please!!!
103453
If $\lim_{x \to a}f(x)=1$ and $\lim_{x\to a}g(x)=\infty$, then  $$\lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x\to a}g(x)(f(x)-1)}.$$ How can one prove this theorem?