Let $b&gt;a&gt;0$ and $f$ continuous. Prove that $$\lim_{\varepsilon\to 0^+}\int_{a\varepsilon}^{b\varepsilon}\dfrac{f(x)}{x}\ \text{d}x=f(0)\ln(b/a).$$ My Try. Solving the integral by parts. Let $u=f(x)$ and $dv=1/x\ dx$. Then, $du=f'(x)dx$ and $v=\ln x$. So, $$\begin{align*} \lim_{\varepsilon\to 0^+}\int_{a\varepsilon}^{b\varepsilon}\dfrac{f(x)}{x}\ \text{d}x&amp;=\lim_{\varepsilon\to 0^+}(f(x)\ln x|_{a\varepsilon}^{b\varepsilon}-\int_{a\varepsilon}^{b\varepsilon}f'(x)\ln x\ \text{d}x)\\ &amp;=\lim_{\varepsilon\to 0^+}(f(b\epsilon)ln(b\epsilon)-f(a\epsilon)\ln(a\epsilon))-\lim_{\epsilon\to 0^+}\int_{a\varepsilon}^{b\varepsilon}f'(x)\ln x\ \text{d}x\\ &amp;=f(0)\ln(b/a)-\lim_{\epsilon\to 0^+}\int_{a\varepsilon}^{b\varepsilon}f'(x)\ln x\ \text{d}x.\end{align*}$$ and there in that last part I got stuck. How can I prove that the integral that remains is 0 or remove it from there?
122687
Let $f:[0, 1] \rightarrow \mathbb{R}$ a continuous function. If $a&gt;0$, show that:  $$ f(0)\ln(\frac{b}{a})=\lim_{\epsilon\rightarrow 0}\int_{\epsilon a}^{\epsilon b} \frac{f(x)}{x}dx$$  Tried using Riemann sum, but did not succeed.
53477
Suppose $\epsilon &gt; 0$ and that $c \in \mathbb{R}$. Since $f^3$ is continuous, we can choose $\delta &gt; 0$ such that whenever $\lvert x - c \rvert &lt; \delta$ we have that $\rvert f^3(x) - f^3(c) \lvert &lt; \epsilon$.   To show that $f$ is continuous, I know I need some number $\delta_1 &gt; 0$ such that $\lvert x - c \rvert &lt; \delta_1$ implies that $\lvert f(x) - f(c) \rvert &lt; \epsilon$.   I've tried manipulating the expression $\lvert f(x) - f(c) \rvert &lt; \epsilon$ by using the fact that $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. I get  $$   \lvert f(x) - f(c) \rvert &lt; \frac{\lvert f^3(x) - f^3(c)\rvert}{\lvert f^2(x) + f(x)f(c) + f^2(c) \rvert} &lt; \frac{\lvert f^3(x) - f^3(c)\rvert}{\lvert f^2(x) + f(x)f(c) \rvert}. $$ However, I am not sure how to continue and I am also not confident that this is a good way to approach the exercise.
83544
If $f(z)$ is continuous in the neighborhood of the point $z=a$, then $$\lim_{r\to 0^+} \int_{\vert z-a\vert =r} \frac{f(z) dz}{z-a}=2\pi i\,f(a)\tag{1}$$  In my desperate intents of proving the above exercise, I tried to prove a 'particular' case of it, which is when $z=0$.   So if $z=0$ and as $f$ is continuous on a neighborhood of zero, given $\epsilon&gt;0$, there exists $\delta&gt;0$ such that $|f(re^{i \theta})-f(0)|\le \epsilon$ for $r\in[0,\delta)$ and $\theta\in\mathbb R$. Hence,  $\left\lvert\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta\right| &lt;2\pi\epsilon \Rightarrow \limsup_{r\to0}\left\lvert\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta\right| \le2\pi\epsilon.$  If $\epsilon\to0$, then $\limsup_{r\to0}\left\lvert\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta\right|=0 \Rightarrow \lim_{r\to0}\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta=0.$  Is this proof correct?  How can I make the proof for $(1)$ following the 'particular proof'? Can someone help me?
69543
If $f$ is continuous, nonnegative on $[a, b]$, show that $\int_{a}^{b} f(x) dx = 0$ iff $f(x) = 0$  "$\Rightarrow$" Assume by contradiction that $f(x) \neq 0$ for some $x_0 \in [a, b]$. Without loss of generality, assume that $f(x_0) &gt; 0$. Since $f$ is continuous, $\exists \delta &gt; 0$ such that $f(x) &gt; 0,  \forall x \in (x_0 - \delta, x_0 + \delta)$. Since $f$ is continuous on $[a, b]$, then it is Riemann integrable. Therefore $\int_{a}^{b} f(x) dx = \sup L(f, p)$ for some partition $p$. Now what I want to do is to show that the lower sum, $L(f, p) &gt; 0$, in that case its supremum is $&gt;0$, and that contradicts the condition that $\int_{a}^{b} f(x) dx = 0$. However, I don't know how to articulate it.  "$\Leftarrow$" If $f(x) = 0, \forall x \in [a, b]$, then $\int_{a}^{b} f(x) dx = \int_{a}^{b} 0 dx = 0$
103524
I have to prove that $f(x)=\log(1+x^2)$ is Uniform continuous in $[0,\infty)$ (with $\epsilon ,\delta$ formulas...)  I wrote the definition: (what I have to prove):  $\forall \epsilon&gt;0 \quad \exists \delta&gt;0 \quad s.t. \quad \forall x,y\in [0,\infty) : \quad |x-y|&lt;\delta \quad \Rightarrow \quad |f(x)-f(y)|&lt;\epsilon $  So I tried developing $|f(x)-f(y)| = |\log(1+x^2)-\log(1+y^2)| = |\log(\frac{1+x^2}{1+y^2})| $...   now this is where I try to make it bigger and simplify the expression so i can choose the right $\delta$ which depends on the $\epsilon$, and then say that if the simplified bigger expression is still smaller than $\epsilon$ then of course the original $|f(x)-f(y)|$ is smaller than $\epsilon$      but what can I do with this expression? any algebraic tricks?
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 $$\Bigg|\int_A^B f(x) dx\Bigg| &lt; \epsilon$$
99861
Let $f$ be a bounded continuous function on $R$. Prove that  $$\lim_{n \to \infty} \frac{n}{\pi} \int_{ R} \frac {f(t)}{1+n^{2}t^{2}} dt=f(0)$$  I solved this question as follows, but I ran into a problem:  Solution:  Since $f$ is continuous at $0$, given $\epsilon&gt;0$, there is $\delta &gt;0$ such that $|f(x)-f(0)| \leq \epsilon, \forall x \in [-\delta \hspace{2 mm} \delta]$. Now consider $f_{n}(x)=\frac{nf(x)}{1+n^{2}x^{2}}$,$n \in N$, is a sequence of continuous function that converges to $0$ on $R-[-\delta \hspace{2 mm} \delta]$ when $n \to \infty$. Also, according to guestion, $f$ is bounded (i.e.$|f_{n}(x)|&lt;g(x)$). Therefore according to Dominated Convergence Theorem   $$\lim_{n \to \infty} \frac{n}{\pi} \int_{ R-[-\delta \hspace{2 mm} \delta]} \frac {f(t)}{1+n^{2}t^{2}} dt=\lim_{n \to \infty} \frac{n}{\pi} \int_{ R-[-\delta \hspace{2 mm} \delta]} 0 \hspace{2 mm} dt=0$$  Now I have a problem to solve the second part. I want to prove     $\lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(t)}{1+n^{2}t^{2}} dt =  \lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(0)}{1+n^{2}t^{2}} dt $  which   $\lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(0)}{1+n^{2}t^{2}} dt =\pi f(0)$  If I can prove that, the rest of question is easy, because I can say:   $\lim_{n \to \infty} n \int_{ R} \frac {f(t)}{1+n^{2}t^{2}} dt=\lim_{n \to \infty} n \int_{ R-[-\delta \hspace{2 mm} \delta]} \frac {f(t)}{1+n^{2}t^{2}} dt+\lim_{n \to \infty} n \int_{ R} \frac {f(t)}{1+n^{2}t^{2}} dt=\lim_{n \to \infty} n \int_{ [-\delta \hspace{2 mm} \delta]} \frac {f(t)}{1+n^{2}t^{2}} dt=\pi f(0)$  Now how can I prove: $\lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(t)}{1+n^{2}t^{2}} dt =  \lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(0)}{1+n^{2}t^{2}} dt $
113738
I am seeking a proof that  $$\int_a^c f(x) dx + \int_c^b f(x) dx  = \int_a^b f(x) dx$$  Please use Stewart's (Calculus: Early Transcendentals, 2016, 8e, p. 378) definition of the definite integral:    This is what I have:  Suppose   $$u=\int_a^c f(x) dx = \lim \limits_{n \to \infty} \sum_{i=0}^n f(x_i^*) (\frac{c-a}{n})$$   and   $$v=\int_c^b f(x) dx = \lim \limits_{n \to \infty} \sum_{i=0}^n f(x_i^*) (\frac{b-c}{n})$$  Then   $$u+v = \lim \limits_{n \to \infty} \sum_{i=0}^n f(x_i^*) (\frac{b-c}{n}) + \lim \limits_{n \to \infty} \sum_{i=0}^n f(x_i^*) (\frac{c-a}{n}) $$  $$=\lim \limits_{n \to \infty} \sum_{i=0}^n f(x_i^*) (\frac{b-c}{n} + \frac{c-a}{n})$$  $$=\lim \limits_{n \to \infty} \sum_{i=0}^n f(x_i^*) (\frac{b-a}{n})$$  $$=\int_a^b f(x) dx$$  Edit: As @Matthew Conroy pointed out, the $f(x_i^*)$ cannot be factored out in the sum because the values of $x_i^*$ are different for each definite integral. So, this proof is invalid.
71040
I am struggling with the proof of the following theorem:  Let $f(x)$ be uniformly continuous in $[a, \infty)$ s.t. the integral $\int_a^{\infty} f(x)dx$ converges. prove that $\lim_{x \to \infty} f(x) = 0$.  I came to the conclusion that it is enough to prove that $\lim_{x \to \infty} f(x)$ exists, and from there I have a proof that the limit is $0$.  I tried to use Cauchy's equivalent definition for the convergence of the improper integral + his definition for a regular limit at $x \to \infty$ with no success...  I will be happy to get hints as for how I should proceed, not proofs, since I really want to crack this one by myself, only I spent a pretty long time with no success.  Thank you :)
30826
Let $f$ a continous function defined in the interval $[0,1]$. Assuming: $\forall x \in [0,1]:f(x) &gt; x$ Prove: $\forall x \in [0,1]:f(x) &gt; x + \varepsilon $  I tried to use  and did some Algebra tricks but it didn't bring me to a safe shore :)    What do you suggest? Thanks!
36912
Theorem. Let $f\colon I \to J$, where $I$ is an interval and $J$ is the image $f(I)$, be a function such that:   $f$ is strictly increasing on $I$. $f$ is continuous on $I$.   Then $J$ is an interval, and $f$ has an inverse function $f^{-1}\colon J\to I$ such that:   $f^{-1}$ is strictly increasing on $J$. $f^{-1}$ is continuous on $J$.   I need to prove only that $f^{-1}$ is continuous on $J$.  Proof. Let $y\in J$, and (for simplicity) assume that $y$ is not an end-point of $J$. Then $y=f(x)$, for some $x\in I$, and we want to prove that $$y_n\to y\implies f^{-1}(y_n)\to f^{-1}(y)=x.$$ Thus we want to deduce that for each $\varepsilon&gt;0$, there is some number $X$ such that $$x-\varepsilon&lt;f^{-1}(y_n)&lt;x+\varepsilon\qquad \forall n&gt;X\qquad(1).$$ Since $f$ is strictly increasing, we know that $$f(x-\varepsilon)&lt;f(x)&lt;f(x+\varepsilon).$$ also, since $y_n\to y=f(x)$, there is some number $X$ such that $$f(x-\varepsilon)&lt;y_n&lt;f(x+\varepsilon), \qquad\forall n&gt;X.$$ Then, applying the strictly increasing function $f^{-1}$ to these inequalities, we obtain (1).  Edit  I don't understand why, since $y_n\to y=f(x)$, there is some number $X$ such that  $$f(x-\varepsilon)&lt;y_n&lt;f(x+\varepsilon), \qquad\forall n&gt;X.$$ I would have written the last line as follows: $$f(x)-\varepsilon&lt;y_n&lt;f(x)+\varepsilon, \qquad\forall n&gt;X.$$ Applying the $f^{-1}$ to my last line, which I think it's correct, I obtain $$f^{-1}[f(x)-\varepsilon]&lt;f^{-1}[y_n]&lt;f^{-1}[f(x)+\varepsilon], \qquad\forall n&gt;X.$$ which is not (1).  What am I misunderstanding? Why the author of this proof wrote that last line and not my version?  Thank you.
88196
Let $f$ be a continuous function on the unit interval $[0,1]$. Show that $$\lim_{n\to\infty}\int_{0}^{1}\cdots\int_0^1\int_{0}^{1}f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)dx_{1}dx_{2}\cdots dx_{n}=f\left(\dfrac{1}{2}\right)$$  This problem is from Selected Problems in Real Analysis. But the author doesn't include a solution. Maybe there is a method for this sort of problem?  Maybe we can use this: $$\int_0^1\cdots\int_{0}^{1}\int_{0}^{1}\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{2}?$$  I also found a similar problem in .
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!
7416
I'd like a hint to solve the following problem:     Given a $f$ continuos and positive on $[a,b]$ prove that $$\lim_{n\to \infty}  \left[\int_a^b (f(x))^n dx\right]^{{1}/{n}}=\sup_{x\in [a,b]} f(x).$$   Thanks a lot!
118621
Task  I have the following function  $f(x)=x^2+1$  I need to prove, according to the $\epsilon - \delta$ definition of a limit, that $f(x)$ is continuous at $x = 2$.  Step 1  $\forall \epsilon &gt; 0 \enspace \exists \delta &gt; 0 : |f(x) - L| &lt; \epsilon, \mbox{ when } 0 &lt; |x - c| &lt; \delta \Rightarrow \lim_{x \to c}f(x)=L$.  If a function is continious at a given point, the value of the limit and the function must be equal, so:  $\lim_{x \to c}f(x)=f(c)$.  And in this case - $f(2) = 2^2 + 1 = 5$.  Which means that $\lim_{x \to c}f(x)=5$.  Step 2  Try to define $x$ in terms of $\delta$.  $0 &lt; |x - 2| &lt; \delta \\ 0 &lt; x &lt; \delta + 2$  Step 3  Try to define $x$ in terms of $\epsilon$.  $ |x^2+1 - 5| &lt; \epsilon \\ 4- \epsilon &lt; x^2 &lt; \epsilon + 4 \\ \sqrt{4- \epsilon} &lt; x &lt; \sqrt{\epsilon + 4} $  Step 4  What next? I'm a little weak on how should I proceed.
63070
$\mathbf{Question}:$ Let $f$ be a continuous function on $[0,1]$. Then prove that the limit $\lim_{n \to \infty} \int_0^1{nx^nf(x)}dx$ is equal to $f(1)$.   $\mathbf{Attempt}$: First, we try to show that the sequence of functions $\{nx^nf(x)\}_{x\in [0,1]}$ is uniformly convergent to $0$ on the restricted domain $[0,1-\epsilon]$, $0&lt;\epsilon&lt;1$.   Let $\sup_{x\in[0,1]}f(x)= \mathcal{M}$. Then $|{nx^nf(x)}|\leq n(1-\epsilon)^n|\mathcal{M}|$ for any $x$ and $n(1-\epsilon)^n \to 0$ as $n\to \infty$.  Thereby, $\lim_{n \to \infty}\int_0^{1-\epsilon}nx^nf(x)dx=\int_0^{1-\epsilon}{\lim_{n \to \infty} }nx^nf(x)dx=0$  Now, denote $\sup_{x\in [1-\epsilon,1]} f(x)=M(\epsilon)$ and $\inf_{x\in[1-\epsilon,1]}f(x)=m(\epsilon)$.  $\int_{1-\epsilon}^1nx^n\ m(\epsilon)dx\leq\int_{1-\epsilon}^{1}nx^nf(x)dx\leq \int_{1-\epsilon}^1 nx^n\ M(\epsilon)dx$.  Now,  $\lim_{n \to \infty}\int_{1-\epsilon}^1 nx^n\ M(\epsilon)dx =\displaystyle \lim_{n\to\infty}\bigg[ \frac{nx^{n+1}}{n+1}M(\epsilon)\bigg]_{1-\epsilon}^1=M(\epsilon)$ [Similarly the other one is $m(\epsilon)$]  Clearly, as $\epsilon \to 0 , \ \ M(\epsilon), m(\epsilon) \to f(1)$, so $\lim_{n \to \infty} \int_0^1{nx^nf(x)}dx=f(1)$.  Is the procedure correct? Kindly verify.
67741
Prove that if $f$ is a continuous real valued function on the interval $[a,b]$ such that $f(x)\ge 0$  for all $x\in [a,b]$ and $f(x)&gt;0$ for some $x\in[a,b]$ then $\int^b_a f(x) \, dx &gt;0$.  The definition I have for Riemann integral is: Let $a,b\in \mathbb{R}$ $a&lt;b$ and let $f$ be a real valued function on $[a,b]$. We say that $f$ is Riemann integrable on $[a,b]$ if there exists a number $A\in \mathbb{R}$ such that for any $\varepsilon&gt;0$ there exists $\delta&gt;0$ such that $|S-A|&lt;\varepsilon$ whenever $S$ is Riemann sum for $f$ corresponding to any partition of $[a,b]$ of width less that $\delta$.