Let $a_n &lt; c &lt; b_n$ and $a_n, b_n → c$ as $n → ∞ $.    Suppose $ f : \mathbb{R} → \mathbb{R}  $ is continuous.     Show that (Possibly using similar proof to FTC...?)  $$  \lim_{n \to\infty }\frac{1}{{b_n - a_n}}\int_{{\,a_n}}^{{\,b_n}}{{f\left(x\right)\,dx}}         = f(c).$$
122435
Let $f \colon [a,b] \to \mathbb{R}$ be a continuous function on $[a,b]$. Prove that there exists $c \in (a,b)$ such that $\frac{1}{b-a}\int_a^b d(s)\,ds = f(c)$.   I understand there is an easy way of doing this, I just want to check if my working to a more complicated method is correct.  $f$ continuous on $[a,b] \implies $ Riemann Integrable on $[a,b]$ $\implies $ makes sense to write $\int_{a}^{b}f(s) ds$  Define $A(x)=\int_{a}^{x}f(s) ds$ then by the Fundamental Theorem of Calculus $A$ is differentiable on $(a,b)$ and $A'(x)=f(x)$ ($f$ is continuous on $[a,b]$)  $A$ differentiable on $(a,b) \implies A$ continuous on $(a,b)$  $\lim\limits_{x \to a+} A(x)=0= A(a) \implies A$ is right continuous at $a$  $\lim\limits_{x \to b-} A(x)=\int_{a}^{b}f(s) ds$ (which exists as $f$ is Riemann Integrable on $[a,b])$$ = A(b) \implies A$ is left continuous at $b$  So $A$ continuous on $[a,b]$ and differentiable on $(a,b)$  By Mean Value Theorem $\exists c \in(a,b)$ st.  $\frac{A(b)-A(a)}{b-a}=A'(c)=f(c) $  (F.T.C) $\implies$  $\frac{\int_{a}^{b}f(s) ds}{b-a}=f(c)$
78587
Let $f:I \rightarrow \mathbb{R}$ derivable in set I. if $\lim x_n = \lim y_n =a \in I$ with $x_n \neq y_n$ and $\lim \frac{f(y_n)-f(x_n)}{y_n - x_n} = f'(a)$ proof that $f'$ is continuous at $a$  my attempt $x_n \neq y_n$ so by MVT $\exists z_n \in I$ like $x_n \geq z_n \geq y_n$ with $\frac{f(y_n)-f(x_n)}{y_n - x_n}=f'(z_n)$ for each $x_n$ and  $y_n$ then $f'(a) = \lim\frac{f(y_n)-f(x_n)}{y_n - x_n}= \lim f'(z_n)$   Using $\lim x_n = \lim y_n = a$ and $x_n \geq z_n \geq y_n \Rightarrow z_n \rightarrow a$  if $ \lim_{x\rightarrow a} f'(x) \neq f'(a)$ then $\exists \epsilon &gt; 0$ such $\forall n\in \mathbb{N}$, $z_n \in [x_n,y_n]$ $0 &lt; |z_n - a| &lt; \frac{1}{n}$ and $|f'(z_n) - f'(a)| \geq \epsilon$. So $z_n \rightarrow a$ but will do not have $\lim f'(z_n) = f'(a)$  so $\lim f'(z_n) = f'(a) = \lim_{x\rightarrow a} f'(x)$ when $z_n \rightarrow a$  and thats implies $f'$ continuous at $a$  my question is, that's correct ? if not, can you help me? thanks
84845
I'm having some difficulty with this proof. Could you help me?     Let $a,b&gt;0$ and $c_n = (a^n + b^n)^{1/n}$. Show that $$\lim_{n\to \infty}c_n = \max \{a,b\}.$$   Thank you.
65492
Let $f$ and $g$ be continuous at $x=a$, with $f(a)&lt; g(a)$. Prove that there exists a neighborhood $I$ of $a\in\mathbb{R}$ such that $f(x)&lt; g(x)$ for all $x \in I$.  It might be helpful to use the theorem that states: If $\lim_{ x\to a} f(x)=L$, $L$ is not $0$, then there exists $\delta&gt;0$ such that $f(x)&gt;L/2&gt;0$ for 0$ &lt; |x-a| &lt; \delta$.
84418
Here's the problem in full. I've stared at it for hours and can't get anywhere, so a hint would be nice.  Suppose that $f:\mathbb N\rightarrow\mathbb R$. If $$\lim_{n\rightarrow\infty}f(n+1)-f(n) = L$$ prove that $\lim_{n\rightarrow\infty}f(n)/n$ exists and equals $L$.  This problem is marked with the word "Cauchy", so I'm guessing Cauchy sequences or something related to them will be necessary, but I can't figure out what.  Edit: It appears as if the Stolz-Cesaro Theorem makes this quite easy. Unfortunately though, we have not discussed this, so I won't be able to use this.
17745
I found this problem on an old FB post and have just been humbled by it.  Let $f(x)$ be continuous on $[0,n]$ such that $n &gt; 0$ and $f(x) + f(n - x)$ does not vanish on $[0,n]$ then evaluate  $$ \int_0^n \frac{f(x)}{f(x) + f(n-x)}dx $$  Any clues/hints would be appreciated.
109939
Let $f$ be a continuous function. I wish to show $$\lim_{n \rightarrow \infty} \int_0^1 n x^n f(x) \; dx = f(1)$$ I can try to split up the integral over intervals $[0, 1-\delta]$, $[1-\delta, 1]$. The integral over $[0, 1-\delta]$ vanishes as $n \rightarrow \infty$, but I am having difficulty estimating the integral over $[1-\delta, 1]$. Thanks!
96948
I must show that if $f$ is continuous on $(0,1)$ and $$\lim_{x \to 0}f(x) = + \infty$$ then $f$ is not uniformly continuous. I am not sure if we have to use epsilon- delta proofs or how to show this.
122905
Let $f:[0,1] \to \mathbb R$ be a continuous function and the sequences $(a_n)_n,(b_n)_n$ s.t. $$\lim_{n\to \infty}  \int_0^1 |f(x)-a_nx-b_n| dx=0.$$ Prove that $(a_n)_n,(b_n)_n$ are convergent.   I know that $$\left|\int_0^1f(x)dx\right| \le \int_0^1|f(x)|dx.$$ Can somebody help me, please?
14373
$$\lim\limits_{n\to\infty} \int_{0}^{1} f(x^n)dx = f(0)$$  when f is continuous on $[0,1]$  I know it can be proved using bounded convergence theorem but,  I wanna know proof using only basic properties of riemann integral and fundamental theorem of calculus and MVT for integrals ...   Thank you.
25113
Suppose $f$ is a continuous function at $x = c$ in $[a,b]$.  Prove that for any sequence ${x_n}$ in $[a,b]$ converting to $c$, the sequence $\{f(x_n)\}$ converges to $f(c)$.  That is, $$ \lim_{n\to\infty}f(x_n)= f\left(\lim_{n\to\infty}x_n\right)$$    This proof seems simple but there are a few things that I need to know first. If $\{x_n\}$ converges to $c$, is it sufficient to substitute $c$ in for $\lim_{n\to\infty}x_n$?  Also needing some guidance on the structure of this proof.  Thanks!
98916
Prove that if $\lim \{a_n\}\to a$ then $\lim \{|a_n|\}\to|a|$. Is the converse true?  I don't know where to start. Should I break this into 2 cases?
27606
If $f$ is continuous in [0,1] then what will be the value of $$\lim_{n\to\infty}\sum_{j=0}^{\left[\frac{n}{2}\right]}\frac{1}{n}f\left(\frac{j}{n}\right)?$$  We know that,$$\lim_{n\to\infty}\sum_{j=0}^{{n}}\frac{1}{n}f\left(\frac{j}{n}\right)=\int_0^1f(x)dx$$But what will be the value if we have $\left[\frac{n}{2}\right]$ instead of $n?$
22355
From Pugh's analysis book, prelim problem 57 from Chapter 4:     Let $f$ and $f_n$ be functions from $\Bbb R$ to $\Bbb R$. Assume that $f_n(x_n)\to f(x)$ as $n\to\infty$ whenever $x_n\to x$. Prove that $f$ is continuous. (Note: the functions $f_n$ are not assumed to be continuous.)   here's my attempt: assume $x_n \to x$. we want to show that $f(x_n) \to f(x)$. so $|f(x_n) - f(x)| \leq |f(x_n)-f_n(x_n)| + |f_n(x_n)-f(x)|$. The second term can be made to be less than any $\varepsilon &gt; 0$ for $n$ sufficiently large. i'm having trouble with the first term. can anyone help? thank you!
114326
Let $f$ be a continuous, nonnegative, real-valued function and $$\int_0^{\infty}f(x)dx&lt;\infty.$$ Prove that $$\lim_{n\to\infty}\frac1{n}\int_0^{n}xf(x)dx=0.$$  A start: If $\lim\limits_{n\to\infty}\int_0^{n}xf(x)dx$ is finite, then it's obvious. Otherwise, perform L'Hopital's rule, we get $\lim nf(n)$, which we want to show is $0$.
80899
Claim: If $f \in {\mathscr R[a,b]}$, then $f$ has infinitely many points of continuity.   1.) I read that it is a corollary of the Lebesgue integrability criterion.  Is it possible to prove the claim without invoking the concept of measure(or using less abstraction) ?    2.) Here is attempt: Given $\epsilon &gt; 0, \exists$ Partition, $P =  \left\{ x_0 =a,...,x_n =b \right\} $ of $[a,b]$ such that $\sum^n_{i=1} (M_i -m_i)\Delta x_i &lt; (b-a)\epsilon$, where  $M_i= \sup \left\{ f(x) : x \in \Delta x_i \right\}$ and $ m_i= \inf \left\{ f(x) : x \in \Delta x_i \right\} $  Let $(M_j -m_j)=\min \left\{ M_i -m_i : i=0,...,n \right\} $.  $ \implies (M_j-m_j)(b-a) \leq \sum^n_{i=1} (M_i -m_i)\Delta x_i &lt;  (b-a)\epsilon$ $ \implies (M_j-m_j)&lt;\epsilon. $  Let $ c\in (x_{j-1},x_j)$ and $\delta$ be any positive number such that $(c-\delta, c+\delta) \subseteq (x_{j-1}, x_j)$.   It follows that $ \left| f(x) - f(c) \right| &lt; \epsilon, $ whenever $ \left| x - c \right| &lt; \delta $.   Since $c$ is arbitrary and $[x_{j-1},x_j]$ is an interval,  there are infinitely many points of continuity.  There is something wrong with my proof since Thomae's function is a counterexample. Could anyone point out the mistakes in my proof? Thank you.
44777
Prove that if $f:(0,\infty)→\mathbb{R}$ satisfying $f(xy)=f(x)+f(y)$, and if $f$ is continuous at $x=1$, then $f$ is continuous for $x&gt;0$.   I let $x=1$ and I find that $f(x)=f(x)+f(1)$ which implies that $f(1)=0$. So, $\lim_{x\to1}f(x)=0$, but how can I use this to prove continuity of $f$ for every $x \in \mathbb R$?   Any help would appreciated. Thanks