$$\lim_{n\to ∞}\frac{1}{n}\log{{n\choose 2\alpha n}}=\frac{3}{2}((1-2\alpha) \log{2\alpha}+2\alpha\log2\alpha)$$ such that $2\alpha n\le n$ I tried to use Stirling formula and we get $$\lim_{n\to ∞}\frac{1}{n}\log{{n\choose 2\alpha n}}=\lim_{n\to ∞}\frac{1}{n}\log\frac{n^{\frac{3n}{2}}}{2\pi(n-2\alpha n)^{\frac{3((n-2\alpha n)}{2}{(2\alpha n)}^{3\alpha n}}}=$$ $$=\lim_{n\to ∞}\log{\frac{n^{\frac{3}{2}}}{2\pi(n-2\alpha n)^{\frac{3((1-2\alpha )}{2}{(2\alpha n)}^{3\alpha }}}}$$ but I couldn’t continue

How can I calculate the following limit: $$ \lim_{n\to \infty} \frac{2\cdot 4 \cdots (2n)}{1\cdot 3 \cdot 5 \cdots (2n-1)} $$ without using the root test or the ratio test for convergence? I have tried finding an upper and lower bounds on this expression, but it gives me nothing since I can’t find bounds that will […]

For a related question, I need to know the $n$th integral of $\ln(x)$ and the fractional derivative of $\ln(x)$. A break down of how fractional derivatives may be found on the Wikipedia. In particular, I need to calculate $\frac{d^{1/2}}{dx^{1/2}}\ln(x)$ and $\frac{d^{-n}}{dx^{-n}}\ln(x)$ where that is the $n$th integral of $\ln(x)$. The fractional derivative in this scenario […]

Please check my proof I will use the fact that $\lim_{x \to c}x^{3}=c^{3}$ equivalent $\lim_{x \to c}x\lim_{x \to c}x\lim_{x \to c}=(c)(c)(c)$ for $\lim_{x \to c} x=c$ we let $\epsilon >0$ and $\delta >0$ we have $0<|x-c|<\delta \leftrightarrow |x-c|<\sqrt[3]{\epsilon }$ but in this case we want to proove $\lim_{x \to c}x^{3}=c^{3}$ We will have $0<|x-c|<\delta \leftrightarrow […]

Compute the limit $$ \lim_{n \rightarrow \infty} \sqrt n \cdot \left[\left(1+\dfrac 1 {n+1}\right)^{n+1}-\left(1+\dfrac 1 {n}\right)^{n}\right]$$ we have a bit complicated solution using Mean value theorem. Looking for others

Qn: If it is given that $$ \lim_{x\to\infty} \frac{x^2 – x – 2}{x + 1} – ax – b = 1 $$ then a and b must be? Now, I tried doing this by 2 methods. Method 1: $$ \frac{x^2 – x – 2}{x + 1} – ax – b $$ $$ = (x – […]

Show that $\lim_{t\rightarrow 1^+}(t-1)\zeta(t)=1$. For $t>1$, we can use the definition $\zeta(t)=\sum_{n=1}^\infty \dfrac{1}{n^t}$, so it is approximately $\int_1^\infty \dfrac{1}{x^t}dx$. How can this lead to the limit?

$\lim\limits_{(x,y) \to (0,0)} f(x,y) = \dfrac{\cos(x) -1 – {x^2 \over 2}}{x^4 + y^4}$ Is the following approach correct? If we approach the origin from $y$ , that is $x = 0$: $\lim\limits_{(x=0,y) \to (0,0)} f(0,y) = {0 \over y^4} = 0$ Now we approach the origin from $x$ and use $\cos(x) \sim_{0} 1 – {x^2\over […]

While doing some tasks for my next calculus course, I ran across this task: “Let $a < b <c$, and assume that $f(x) \le g(x)$ for all $x \in [a, c]$. If $\lim_{x \to b}f(x) = L$ and $\lim_{x \to b}g(x) = M$, prove that $L \le M$.” I have done some thinking about this, […]

I know to how prove normal limits using the epsilon-delta definition, say: $$\lim_{x\to a}f(x) = L$$ But, there was a question on my textbook which I couldn’t quite figure out to do, even though I’ve thought about it for a while I don’t even know how to go about starting it. Use $\varepsilon$-$\delta$ definition of […]

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