Let $f(x)$ be a monotonic increasing function on $[0,\infty)$ and for every $x>0$ it is integrable in $[0,x]$, so that $\lim_{x\to \infty}\frac{1}{x}\int_{0}^{x}f(t)dt=a$. I need to prove that $\lim_{x\to \infty}f(x)=a$. I tried to use the limit definition: $|\frac{1}{x}\int_{0}^{x}f(t)dt -a|<\epsilon$ and to use the fact that $\int_{0}^{x}f(t)dt\geq x\inf f(x)$, but I can’t see how does it help […]

I’m a bit confused with the general concept of convergence of a sequence of sets. I’m well aware that the limit of a sequence $\{C^{\nu}\}$ exists iff $$\liminf_{\nu \rightarrow \infty} C^{\nu} = \limsup_{\nu \rightarrow \infty} C^{\nu}$$ where lim inf (resp. lim sup) is the set of points that appear in the limit all but finitely […]

Undoubtedly quite simple, but I’m stuck: If $f$: $[a, b)\longrightarrow\mathbb{R}$ is uniformly continuous, must $\lim\limits_{x\longrightarrow b}f(x)$ exist?

This question is about where I made my mistake in the computation of a limit. It relates to An answer I gave that confused me. The question to which I gave the (partial) answer is related to tetration but my mistake is probably a simple general one ( considering tetration as complicated ). Here is […]

Inspired by this question I’m trying to prove that $$\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!} \approx e^{\frac{x}{2}}$$ So I needed to find the value of $$\frac{\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!}}{e^{\frac{x}{2}}} = \frac{\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!}}{\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{\frac{x}{2}^k}{k!}} \\ = \lim_{m \to \infty} […]

here’s the question, how can I solve this: $$\lim_{x \rightarrow \infty} x\sin (1/x) $$ Now, from textbooks I know it is possible to use the following substitution $x=1/t$, then, the ecuation is reformed in the following way $$\frac{\sin t}{t}$$ then, and this is what I really can´t understand, textbook suggest find the limit as $t\to0^+$ […]

I have rewritten this entire question, since what I’ve learned since asking it requires me to restate it. I want to get rid of the obfuscating revisions. Let’s say that f is a continuous function. $f(x)$ approaches L as x approaches a. So $\lim\limits_{x \to a}f(x) = L$ When it’s said that the gradient of […]

Suppose I have two sequences of random variables $\{x_n\}$ and $\{y_n\}$ such that $x_n\leq y_n$ and $\text{plim}\;x_n=L_x$ and $\text{plim}\;y_n=L_y$, can I say $L_x\leq L_y$ (almost surely)? Does it matter if I further impose that $L_x$ and $L_y$ are nonrandom? I tried to replicate the argument for the nonstochastic case (included below for completeness) but I […]

If $0 \lt x \lt 1$ and $$A_n=\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+…..+\frac{x^{2^n}}{1-x^{2^{n+1}}}$$ then Find $\lim\limits_{n\to \infty}A_n$.

How to calculate this question? $$\lim\limits_{{\rho}\rightarrow 0^+}\frac{\log{(1-(a^{-\rho}+b^{-\rho}-(ab)^{-\rho}))}}{\log{\rho}} ,$$ where $a>1$ and $b>1$. Thank you everyone.

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