Been struggling with this for a while now, just kind of spinning my wheels at this point.  Any help/advice/anything much appreciated.  This is a problem from Wade's Intro to Analysis, chapter 3.  Suppose $f : \textbf{N} \to \textbf{R}$. If $\lim\limits_{n \to \infty}f(n + 1) - f(n) = L$, prove that $\lim\limits_{n \to \infty}\frac{f(n)}{n}$ exists and equals $L$.
84418
Here's the problem in full. I've stared at it for hours and can't get anywhere, so a hint would be nice.  Suppose that $f:\mathbb N\rightarrow\mathbb R$. If $$\lim_{n\rightarrow\infty}f(n+1)-f(n) = L$$ prove that $\lim_{n\rightarrow\infty}f(n)/n$ exists and equals $L$.  This problem is marked with the word "Cauchy", so I'm guessing Cauchy sequences or something related to them will be necessary, but I can't figure out what.  Edit: It appears as if the Stolz-Cesaro Theorem makes this quite easy. Unfortunately though, we have not discussed this, so I won't be able to use this.
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 $$\Bigg|\int_A^B f(x) dx\Bigg| &lt; \epsilon$$
14823
Prove: if $f''(x_0)$ exists then $\lim\limits_{x \rightarrow x_0} \dfrac{f(x_0+h)-2f(x_0)+f(x_0-h)}{h^2} = f''(x_0)$.    I'm not exactly sure how Taylor's theorem fits into all this, but I found the first derivative, and I haven't been able to progress past that.
18978
Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$  Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?  $\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| &lt; \epsilon$  $\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| &lt; \epsilon$  $\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} &lt; \epsilon$  $\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} &gt; \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$  $\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow \sqrt{n} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow n &gt; \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
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?
100496
Not sure if my proof as is is correct:  Since $a_n \leq b_n$ for all $n$, we have $a_n - b_n \leq 0$. As such, by the limit location theorem it holds that $\lim_{n\to\infty} (a_n - b_n) \leq 0$. By linearity, we have $\lim_{n\to\infty} a_n - \lim_{n\to\infty} b_n \leq 0$, which implies $lim_{n\to\infty} a_n \leq \lim_{n\to\infty} b_n$.  My one concern is using the LL Theorem on the sequence $a_n - b_n$: do I need to prove this sequence is convergent first?
76626
Prove that:  {$f_n$} converges unifromly to $f$ on $E$ as $n \rightarrow \infty$ if and only if $\lim_{n\rightarrow \infty} ||f_n-f||_u = 0$  I don't understand the approach for the Forward proof.  My approach is:  By definition, for $\epsilon &gt; 0$, there exist $N \in \mathbb{N}$ such that $|f_n(x) - f(x)| &lt; \epsilon$; hence, the smallest upper bound should also be less than the $\epsilon$ (i.e., ${sup}_{x \in E}|f_n(x)-f(x)| &lt; \epsilon$). Thus, $\lim_{n\rightarrow \infty} ||f_n-f||_u = 0$.  However, the solution uses $\frac{\epsilon}{2}$ as the following:     $|f_n(x)-f| \leq {sup}_{x \in E}|f_n(x)-f(x)| &lt; \epsilon $      ${sup}_{x \in E}|f_n(x)-f(x)| \leq \frac{\epsilon}{2}$      ${sup}_{x \in E}|f_n(x)-f(x)| \leq \frac{\epsilon}{2} &lt; \epsilon$   Hence, $\lim_{n\rightarrow \infty} ||f_n-f||_u = 0$.  Why?
31983
If $-1&lt;x&lt;1$ show that $\lim\limits_{n \to \infty}{n\,x^n} = 0$. I don't have idea. I only prove that $n\,x^n$ is decreasing.
37682
Today my lecturer put up on the board that:  If $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}$ exists and $a_n&gt;0$ then   $\displaystyle \limsup\limits_{n\to\infty}\left(a_n^{\frac{1}{n}}\right)=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$  however I am not sure why this is true, can somebody give me a hint or something as to how to go about proving this.  thanks for any help
68731
$f\colon [0,\infty) \to \mathbb{R}$ is uniformly continuous.  If $\lim_{n\to \infty}f(n+x)=0$ where $x\in [0,1]$, then $\lim_{x\to\infty} f(x)=0$.  every my trying to solve this is wrong :(  how can i prove this?  and if $f$ is just continuous, what happens to the results
114069
$U(n)$ and $V(n)$ are sequences. $U(n)&gt; 0$  and   $V(n) &gt;0$ such that: ${V(n+1)\over V(n)}≤ {U(n+1)\over U(n)}$.   Prove that:   If $\,\,\lim\limits_{n\to\infty} V(n)= +\infty\,\,$ then $\,\,\lim\limits_{n\to\infty}  U(n) = +\infty$ If $\,\,\lim\limits_{n\to\infty} U(n) = 0 \,\, $  then  $\,\,\lim\limits_{n\to\infty} V(n) = 0$   Can anyone help ?
111937
I'm doing exercise in Real Analysis of Folland, and got stuck on this problem. I try to use Fatou lemma but can't come to the conclusion. Can anyone help me. I really appreciate.          Consider a measure space $(X, M, \mu)$.Suppose $\{f_{n}\} \in L^{+}, f_{n} \rightarrow f$ pointwise and $\int f = \lim \int f_{n} &lt; \infty$. Then $\int_{E} f = \lim \int_{E} f_{n}$ for all $E \in M$. However, this need not be true if $\int f = \lim \int f_{n} = \infty$      Thanks so much.
93025
Alright, so I've already proven that both $\forall n \in \mathbb{N}:\lim_{n\rightarrow\infty}\sqrt[n]{n} = 1$ and $\forall K\geq 1:\lim_{n\rightarrow\infty}\sqrt[n]{K}=1$.   I got the feeling, that I can prove $\forall  K \in \mathbb{R}&gt; 0: \lim_{n\rightarrow\infty} \sqrt[n]{K} = 1$ with a simple limit comparison test but I can't figure out how exactly.
7014
Hello I was hoping that someone might be able to verify that the following proof that $\lim_{x\to 0} {1\over x}$ does not exist is correct.   First assume that $\lim_{x\to 0} {1\over x} = L$. This means that for every $\epsilon &gt; 0$ there is a $\delta &gt; 0$ such that for all $x$ if $0 &lt; \lvert x\rvert &lt; \delta$ then $\lvert{1\over x} - L \rvert &lt;\epsilon.$  Now let $\epsilon=1$ and choose $x &lt; \min(1,\delta)$.  Therefore there is a $\delta&gt;0$ such that if $0 &lt; \lvert x\rvert &lt; \delta$ then $\lvert{1\over x} - L \rvert &lt;\epsilon$.  It follows that $\lvert{1\over x}\rvert &lt; 1+ \lvert L\rvert$.  However clearly $\lvert{1\over x}\rvert&gt;1$ and therefore a contradiction has been reached and there is no number $L$ such that $\lim_{x\to 0} {1\over x} = L$.
45476
I'm trying to solve this question:     Let $f:[0,+\infty) \to \mathbb{R}$ be derivable and $\lim_{x \to +\infty} f'(x) = L$, then $\lim_{x\to \infty}\frac {f(x)}{x}=L$.   I'm trying to solve this question using l'Hôpital rule, but I couldn't use it, because I don't know if $\lim_{x\to +\infty} f(x)=+\infty$.  I need help.  Thanks a lot.
24990
Let $(x_n)_{n \ge 1}$ be a sequence of real numbers satisfying  $$x_{m+n} \le x_n+x_m$$  $m,n \ge 1$. Show that $\lim \limits_{n \to \infty} \dfrac{x_n}{n}$ exists and is equal to $\inf  \left \{\dfrac{x_n}{n}; n\ge 1 \right \}$.
58710
What would be the nicest proof of the following theorem:  If $\lim\limits_{n \rightarrow \infty} a_{n} = \infty$, then $\lim\limits_{n \rightarrow \infty} \left(1 + \frac{1}{a_{n}} \right) ^ {a_{n} } = e$. If $\lim\limits_{n \rightarrow \infty}b_{n} = 0$, then $\lim\limits_{n \rightarrow \infty} \left(1 + b_{n} \right) ^ {\frac {1} {b_{n}} } = e$.  I somehow failed to find a proof here on the website and in the literature.