I've got some inquiries while doing exercise. I want to share it and ask advices.  Here's the problem.    We have uniformly continuous function $f:[0, \infty)$ $\to$ $\mathbb R$ satisfies $\lim\limits_{n\to\infty}$ $f(n+x)=0$ whenever $x \in [0, 1]$ and $n \in \mathbb N$. Show that $\lim\limits_{x\to\infty}$ $f(x)=0$. And if the uniformly continuous condition changes into continuous condition, the statement still holds?    In my trial, I approached with the $\epsilon$ - $\delta$ method and made some outcome but little bit unsatisfied. And I have difficulty in finding some easy counter-example of the if-question. I wonder there are any ideas better than $\epsilon$ -$\delta$ method and easy counter-examples.  EDIT:  Here's the specification about the curiosity I have.  In this problem, the difference between limit through natural number and through real number matters. I think because of $\mathbb N$ $\subsetneq$ $\mathbb R$, limit through $\mathbb R$ implies limit through $\mathbb N$ but the converse doesn't. What I had to request was the logical explanation about this difference and background understanding of this theme. This situation is very common in analysis but I haven't seriously thought about it.
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$f\colon [0,\infty) \to \mathbb{R}$ 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
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Let $f(x) = \frac{1}{x}$ for $x &gt; 0$ and take our set at which the function act on $(0,1]$. This function is continuous but not uniformly continuous on $A$.  To prove this consider $\epsilon = 10$,and suppose we have $\delta$, $0 &lt; \delta &lt; 1$. Taking $x = \delta$ and $p = \frac{\delta}{11}$, we get  $|x - p| &lt; \delta$ and $|f(x) - f(p)| = \frac{11}{\delta} - \frac{1}{\delta} &gt; 10$.   In the following example from my book shows this is continuous but not uniform continuous but the example is missing correct ? since we also need to find counter example for $\delta \geq 1$ right ? If someone could present that example too that would be nice.  Also my question is different since the possible duplicate above shows very different question to the question I am asking here since I am asking why we must consider $\delta \geq 1$ so please read carefully my question.
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Suppose that $f:\Bbb R\to \Bbb R$ is continuous and that $f(x)\to 0$ as $x\to \pm\infty$. Prove that $f$ is uniformly continuous.   Since $f(x)\to 0$ as $x\to \pm\infty$ we have $|f(x)|&lt;\epsilon $ whenever $x&lt;-K$ and $x&gt;K$ where $K&gt;0$.  Also $f$ is uniformly continuous on $[-K,K]$ as the domain is compact.  Hence $f$ is uniformly continuous in $(-\infty,-K)\cup [-K,K]\cup (K,\infty)$.  But how to conclude that $f$ is uniformly continuous on $\Bbb R$ from above as I may chose $x\in (-\infty,-K);y\in [-K,K]$ ,how to conclude that $|f(x)-f(y)|&lt;\epsilon $ from above for a given $\epsilon$?  Please suggest the edits required.
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Using the definition of continuity and the $\epsilon$-$\delta$ definition of a limit, how would you proceed to show that the following statement holds true?  Let $f : \Bbb{R} \rightarrow \Bbb{R}$ be a continuos function at $x=0$, with the property that if $x\neq 0$ then $f(x)\geq 0$. Then $f(0) \geq 0$.  My attempt so far has been to assume that $f(0) \lt 0$, which means that $$f(0)=\lim_{x\to 0} f(x)\lt 0,$$ because of the continuity at the given point. Furthermore, I think this means that when $x$ approaches $0$ from both sides, at some point the value $f(x)\lt 0$, thus forming a contradiction.   I can't seem to see what to do afterwards. Is this a way of approaching the problem? And how will the $\epsilon-\delta$ definition come into play if my approach is usable?
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Another question I am struggling with:  Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be defined in a neighborhood of $a$ (except maybe $a$ itself) and suppose that, $\lim_{x \rightarrow a}\left(f(x)+\dfrac{1}{\left|f(x)\right|}\right)=0$. Find $\lim_{x \rightarrow a}f(x)$, and prove by definition, that this is indeed the limit.    Thank you all for your answers!!  But.. How can I prove by definition that this is indeed the limit? take $\epsilon&gt;0$, I have to show that there exists a $\delta&gt;0$ such that, $|x-a| \leq \delta$ implies $|f(x)+1| \leq \epsilon$. now, I do know that there exists an $\delta_1&gt;0$, such that $|f(x)+\frac{1}{|f(x)|}|&lt; \epsilon$.Or, $\frac{\epsilon}{k}$ for some constant $k$ But, how can I continue from there?
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Put precisely, if $f:\mathbb R\to\mathbb R$ is continuous with  $$ \lim_{s\to 0^+}s^{-1}(f(t+s)-f(t-s))=0 \kern20mm (*) $$ for all $t\in\mathbb R$ , is $f$ then constant?  I guess it is but I cannot see the proof. Obviously, if instead of ($*$) we had $\lim_{s\to 0^+}s^{-1}(f(t+s)-f(t))=0$ , i.e. the right-handed derivative zero, then the assertion could be proved indirectly by a simple compactness argument applied to closed intervals $[a,b]$ with $f(a)\not=f(b)$ . However, I cannot see how this argument could be modified to get the above result. Is there possibly a reference to this, or is the result even false?
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Let $f\colon(0,\infty)\to \mathbb R$ be a continuous function such that $f(x)\le f(nx) , \forall n \in \mathbb N , \forall x &gt;0$ , then is it true that $\lim_{x\to \infty} f(x)$ exists (may be infinite) ?
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Let $f_n(x) : \mathbb{R} \rightarrow \mathbb{R}$ be defined as $\dfrac{x}{n}$. Is $f_n$ uniformly convergent?   With "$(a_n)$ is limited and $(b_n)$ converges to 0 $\Rightarrow (a_nb_n)$ converges also to 0" i would guess that $x\cdot \dfrac{1}{n}$ converges uniformly but $x$ is not limited in this case. So does ist converge uniformly or not?
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Let $l \in\mathbb{R} $ and $ f:(0,\infty) \to \Bbb R $ be a function such that $ \lim_{x\to \infty} xf(x)=l $ .  Prove that  $ \lim_{x\to \infty} f(x)=0 $.  Any help would be appreciated - I found this difficult to prove
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Is $f(x)=\ln(x)$ uniformly continuous on $(1,+\infty)$? If so, how to show it?  I know how to show that it is not uniformly continuous on $(0,1)$, by taking $x=\frac{1}{\exp(n)}$ and $y = \frac{1}{\exp(n+1)}$.  Also, on which interval does $\ln(x)$ satisfy the Lipschitz condition?
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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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My Approach. To prove the given statement it is sufficient to show that $$\exists \epsilon &gt; 0, \exists x \in [0,1], \forall N \in \mathbb{N}, \exists n &gt; N $$ $$ \Rightarrow | f_n(x) - f(x)| &gt; \epsilon$$    Let $\epsilon = \frac{1}{10}, x = \frac{1}{n}, n =N+1$. Hence $| f_n(x) - f(x)| = \frac{1}{2} &gt; \epsilon = \frac{1}{10}$. Hence, we have shown for every $N \in \mathbb{N}$, how to choose $n$ such that $| f_n(x) - f(x)| &gt; \epsilon$. Thus, $f_n(x)$ does not converge uniformly on $[0,1]$.    Is there any logical flaw or wrong steps in my proof. Any help would be appreciated.
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Let $f\in C^{\infty} ([-1,1])$ and suppose there's a constant $M&gt;0$ such that $|f^{(j)}(x)|\le M:\forall j\in\mathbb N$ (including zero) and for all $x\in [-1,1]$. Prove that if $f(\frac 1 k)=0 :\forall k\in \mathbb N$ then $f(x)=0 :\forall x\in [-1,1]$  I understand that the function is always smooth and Lipschitz, so it's always uniformly continuous too. So it can't be that $f(\frac 1 k)=0$ and for the closest irrational number say $y$ we will have $f(y)\neq0$ because that will contradict uniform continuity. It's a bit similar to Dirichlet function. If I'm right then it can be proved without using the fact that the function is smooth and I should be able to find for $f^{(0)}$ a contradiction which should be enough but it doesn't feel right if I'm not using parts of the question. Also, I'm not sure on how to formally express the counter example, should i use the $\epsilon - \delta$ ? Heine with two sequences ?
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Let $f:\mathbb R \to \mathbb R$. Prove that if $f$ is uniformly continuous on $[a,b]$ and $[b,c]$, then $f$ is uniformly continuous on $[a,c]$.  My attempt at a solution:  I've came up with a solution but I am having doubts if it is correct, so I would like to check that:  Let $\epsilon&gt;0$, we know that there exist $\delta_1, \delta_2$ such that  if $x,y \in [a,b]  |x-y|&lt;\delta_1 \implies |f(x)-f(y)|&lt;\dfrac{\epsilon}{2}$   and if   $x,y \in [b,c], |x-y|&lt;\delta_2 \implies |f(x)-f(y)|&lt;\dfrac{\epsilon}{2}$  Let $\delta=\min\{2\delta_1,2\delta_2\}$, if $|x-y|=|x-b+b-y|\leq |x-b|+|b-y|&lt;\delta \implies |f(x)-f(y)|\leq |f(x)-f(b)|+|f(b)-f(y)|\leq \dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon$. This proves that $f$ is uniformly continuous.   I would appreciate if anyone could tell me if my proof is ok or if I've made any mistakes. (S.A. Understanding Analysis. pp 119 question 4.4.7)
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The question is the following: Assume $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous, and for all $x \neq 0$, $f'(x)$ exists. If $\lim_{x \rightarrow 0} f'(x) = L$ exists, does it follow that $f'(0)$ exists?  My intuition is that this does not have to hold in general. However, I keep finding counter examples such that $f'(0)$ exists where $\lim_{x \rightarrow 0} f'(x) = L$ does not exist, which is obviously not the original question.
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It seems intuitively very clear that $e^{x}$ is not uniformly continuous on $\mathbb{R}$. I'm looking to 'prove' it using $\epsilon$-$\delta$ analysis though. I reason as follows:  Suppose $\epsilon &gt; 0$; in fact, fix it to be $\epsilon=1$.   For contradiction, suppose that $\exists \delta &gt;0$ s.t. $$ (\star) \ |x-y|&lt;\delta \Rightarrow |e^{x}-e^{y}|&lt;\epsilon=1 \text{  for  } x,y \in \mathbb{R}.$$ Note that $e^{x+\delta}-e^{x}=e^{x}(e^{\delta}-1)$. So, for $x$ large enough (so that RHS $&gt;1$), the relation $(\star)$ does not hold.  This is our contradiction, and so the exponential function is not uniformly continuous on $\mathbb{R}$.  Is this reasoning correct and sufficient?  Thanks.
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Be $f:\mathbb R^ +\mapsto\mathbb R$ a function that satisfies the following conditions:  a)$ f(f(f(x)))+2x=f(3x)$ for every $x\gt 0$;  b)  $\lim_{x \to \infty} (f(x)-x)=0$.  This was proposed by Gabriel Dospinescu  at a romanian competitions about a decade ago. No idea how to approach it. Thank you!