Articles of limits

Limit of $\lim_{t \to \infty} \frac{ \int_0^\infty \cos(x t) e^{-x^k}dx}{\int_0^\infty \cos(x t) e^{-x^p}dx}$

Let \begin{align} f(t,k,p)= \frac{ \int_0^\infty \cos(x t) e^{-x^k}dx}{\int_0^\infty \cos(x t) e^{-x^p}dx}, \end{align} My question: How to find the following limit of the function $f(t,k,p)$ \begin{align} \lim_{t \to \infty} f(t,k,p), \end{align} for any $p>0$ and $k>0$. What is known Some facts about the function Note that $\int_0^\infty \cos(x t) e^{-x^k}dx$ is a fourierier transform of $e^{-{|x|^k}}$. […]

Question about the derivative definition

The derivative at a point $x$ is defined as: $\lim\limits_{h\to0} \frac{f(x+h) – f(x)}h$ But if $h\to0$, wouldn’t that mean: $\frac{f(x+0) – f(x)}0 = \frac0{0}$ which is undefined?

Computing $\lim\limits_{n\to\infty} \int_{0}^{\infty}\frac{\sin(x/n)}{(1+x/n)^{n}}\, dx$

$$\lim_{n\to\infty} \int_{0}^{\infty}\frac{\sin(x/n)}{(1+x/n)^{n}}\, dx$$ I’ve been able to show that the integral is bounded above by 1 (several ways). One of the simplest is just letting $u=x/n$ and doing some bounding above. It must converge to something (if it is monotone increasing.. don’t think so), but all my efforts to use Lebesgue Dominated Convergence have failed. […]

If $x_n\leq y_n$ then $\lim x_n\leq \lim y_n$

This question already has an answer here: Suppose that $(s_n)$ converges to $s$, $(t_n)$ converges to $t$, and $s_n \leq t_n \: \forall \: n$. Prove that $s \leq t$. 3 answers

Compute $\sum_{k=1}^{\infty}e^{-\pi k^2}\left(\pi k^2-\frac{1}{4}\right)$

How may I evaluate the below series? $$\sum_{k=1}^{\infty}e^{-\pi k^2}\left(\pi k^2-\frac{1}{4}\right)$$ I’m supposed to come up with a solution by only using high school knowledge. Thanks in advance for your hints, suggestions!

Proof that $\lim_{n\rightarrow \infty} \sqrt{n}=1$

Thomson et al. provide a proof that $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$ in this book. It has to do with using an inequality that relies on the binomial theorem. I tried to do an alternate proof now that I know (from elsewhere) the following: \begin{align} \lim_{n\rightarrow \infty} \frac{ \log n}{n} = 0 \end{align} Then using this, I […]

Finding the limit of $\left(\frac{n}{n+1}\right)^n$

Find the limit of: $$\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n$$ I’m pretty sure it goes to zero since $(n+1)^n > n^n$ but when I input large numbers it goes to $0.36$. Also, when factoring: $$n^{1/n}\left(\frac{1}{1+\frac1n}\right)^n$$ it looks like it goes to $1$. So how can I find this limit?

How to prove that $\lim_{n \to \infty} n x^{n} = 0 $ when $0<x<1$?

Intuitively it’s easy, but hard to prove by the epsilon-delta method: $$ \lim_{n \to \infty} n x^{n} = 0$$

Limit of $\prod\limits_{k=2}^n\frac{k^3-1}{k^3+1}$

Calculate $$\lim_{n \to \infty} \frac{2^3-1}{2^3+1}\times \frac{3^3-1}{3^3+1}\times \cdots \times\frac{n^3-1}{n^3+1}$$ No idea how to even start.

How to find this limit: $A=\lim_{n\to \infty}\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\cdots+\sqrt{\frac{1}{n}}}}}$

Question: Show that $$A=\lim_{n\to \infty}\sqrt{1+\sqrt{\dfrac{1}{2}+\sqrt{\dfrac{1}{3}+\cdots+\sqrt{\dfrac{1}{n}}}}}$$ exists, and find the best estimate limit $A$. It is easy to show that $$\sqrt{1+\sqrt{\dfrac{1}{2}+\sqrt{\dfrac{1}{3}+\cdots+\sqrt{\dfrac{1}{n}}}}}\le\sqrt{1+\sqrt{1+\sqrt{1+\cdots+\sqrt{1}}}}$$ and it is well known that this limit $$\sqrt{1+\sqrt{1+\sqrt{1+\cdots+\sqrt{1}}}}$$ exists. So $$A=\lim_{n\to \infty}\sqrt{1+\sqrt{\dfrac{1}{2}+\sqrt{\dfrac{1}{3}+\cdots+\sqrt{\dfrac{1}{n}}}}}$$ But can use some math methods to find an approximation to this $A$ by hand? and I guess maybe this is true: […]