Exercise 2.5 of Izenman’s Modern Multivariate Statistical Techniques: Consider a hypercube of dimension $r$ and sides of length $2A$ and inscribe in it an $r$-dimensional sphere of radius $A$. Find the proportion of the volume of the hypercube that is inside the hypersphere, and show that the proportion tends to $0$ as the dimensionality $r$ […]

The question is this. In $h(x) = \dfrac{\sqrt{x+9}-3}{x}$, show that $\displaystyle \lim_{x \to 0} \ h(x) = \frac{1}{6}$, but that $h(0)$ is undefinied. In my opinion if I use this expression $\displaystyle \lim_{x \to 0} \dfrac{\sqrt{x+9}-3}{x}$ above with the $-3$ inside the square root I got an undefinied expression, but if I put the $-3$ […]

To my knowledge there are two possible ways to define $e^x$ $$e^x = \sum_{i=0}^{\infty}\frac{x^i}{i!}$$ $$e^x = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n$$ So my question is: Why does… $$\sum_{i=0}^{\infty}\frac{x^i}{i!} = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n$$

I have read this solution, but I could not understand it. It has shown $$\sigma_n\leqslant \frac 1n\sum_{j=1}^ks_j+\sup_{l\geqslant k}s_l,$$ but how does it go to $$\sup(\sigma_n)\leqslant \frac 1n\sum_{j=1}^ks_j+\sup_{l\geqslant k}s_l$$ I have thought of regarding the RHS as upper bound of $\sigma_n$, but I am not sure about this, since RHS varies as $n$ changes. The writer […]

Working on convergence and divergence of infinite series, I recently focused my attention on the summation $$\displaystyle\sum\limits_{n=2}^{\infty} \frac{1}{log^n(n)}$$ While proving the convergence of this series is trivial (e.g., using the root test), finding a closed-form expression for the value to which it converges seems to be hard. The summation converges to $ \displaystyle\approx 3.24261$. After […]

Find $\lim_\limits{x\to 0}{x\left[1\over x\right]}$. $Attempt$: $\lim_\limits{x\to 0}{x\left[1\over x\right]}=\lim_\limits{x\to 0}x\left ({1\over x}-\{{1\over x}\}\right)=\lim_\limits{x\to 0}\left (1-{x\{{1\over x}\}}\right).$ Since $\{{1\over x}\}$ is bounded and $x\to 0$, then $x\cdot\{{1\over x}\}\to 0$ and therefore $\lim_\limits{x\to 0}{x\left[1\over x\right]}=1.$

In order to find the slope of the tangent line at the point $(4,2)$ belong to the function $\frac{8}{\sqrt{4+3x}}$, I choose the derivative at a given point formula. $\begin{align*} \lim_{x \to 4} \frac{f(x)-f(4)}{x-4} &= \lim_{x \mapsto 4} \frac{1}{x-4} \cdot \left (\frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{4+3 \cdot 4}} \right ) \\ \\ & = \lim_{x \to 4} \frac{1}{x-4} \cdot \left […]

Why is $\displaystyle \lim_{x \to \infty} \ x^{2/x} = 1$ since this is an indeterminate form $\infty^{0}$ and I can’t see any manipulation that would suggest this result?

Assume ${a_n}$ is a positive, strictly increasing sequence, while $a_{n+1}-a_n$ is bounded. Prove: for every real number $\alpha \in (0,1)$, $$\lim_{n \to \infty}(a_{n+1}^{\alpha}-a_n^{\alpha})=0.$$ I’m wondering how can I associate $a_{n+1}-a_n$ with $a_{n+1}^{\alpha}-a_n^{\alpha}$.

I have a question concerning a special limit $$\lim_{n \rightarrow \infty}\frac{a_n}{b_n}=1$$ Can I always conclude from this that $$\lim_{n \rightarrow \infty}{(a_n-b_n)}=0$$ Even if $a_n$ and $b_n$ are divergent? I tried to find some counter examples but I couldn’t.

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