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Question

Me gustaría calcular %#% $ #%

hemos %#% $ #%

Tenga en cuenta que:

$$\lim\limits_{x\to a}\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)},\quad a \in\mathbb{R}^* \,\,\text{fixed} $$

Ahora tenemos $$\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)}=e^{\tan\left( \dfrac{\pi x}{2a}\right)\ln\left(2-\dfrac{x}{a} \right)}.$, i.e.$$\ln\left(2-\dfrac{x}{a} \right)\sim_{a}1-\dfrac{x}{2}.$ y $\dfrac{\pi x}{2a}\underset{x\to a}{\longrightarrow} \dfrac{\pi}{2}$

Estoy atrapado.

Actualización: aquí está otra manera:

\begin{aligned} \left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)}&=\exp\left[{\tan\left( \dfrac{\pi x}{2a}\right)\ln\left(2-\dfrac{x}{a} \right)}\right].\\ &=\exp\left[ \left(1-\dfrac{x}{a}\right)\tan\left(\dfrac{\pi x}{2a}-\dfrac{\pi}{2}+\dfrac{\pi}{2} \right).\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{1-\dfrac{x}{a}}\right]\\ &=\exp\left[ -\dfrac{\left(1-\dfrac{x}{a} \right)}{\tan\left(\dfrac{\pi x}{2a}-\dfrac{\pi}{2} \right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\ &=\exp\left[ -\dfrac{\left(1-\dfrac{x}{a} \right)}{\tan\left(-\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)\right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\ &=\exp\left[ \dfrac{2}{\pi} \dfrac{\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)}{\tan\left(\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)\right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\ \end{alineado}

Así $\dfrac{\pi x}{2a}-\dfrac{\pi}{2} \underset{x\to a}{\longrightarrow} 0$ $

• Soy que soy justo al lado de lo interesado de manera que usar equivalentes

Answer 1

Sugerencia:

$$\left(2-\frac xa\right)^{\tan(\pi x/2a)}=\left[\left(1+\left(1-\frac xa\right)\right)^{1/(1-x/a)}\right]^{(1-x/a)\tan(\pi x/2a)}\\\stackrel{x\to a}\longrightarrow e^{2/\pi}$$

Ya que,

$$e=\lim_{u\to0}(1+u)^{1/u}$$

$$\frac2\pi=\lim_{u\to1}(1-u)\tan(\pi u/2)$$

para aquellos que quieren hacerlo riguroso, utilice Teorema del apretón:

$$\underbrace{\left(2-\frac xa\right)^{\frac2\pi\left(\frac a{a-x}-1\right)}}_{\text{Laurent expansion of }\tan(z)\text{ at }z=\frac\pi2}\le\left(2-\frac xa\right)^{\tan(\pi x/2a)}\le\underbrace{e^{\left(1-\frac xa\right)\tan\left(\frac{\pi x}{2a}\right)}}_{\text{Maclaurin expansion of }e^x}$$

donde las desigualdades son verdaderas $x<a$ y volteado para $x>a$.

Answer 2

Podemos también aplicar a L'Hopitals a $\log\left(2-\frac{x}{a}\right)\tan \frac{\pi x}{2a}$ y luego apelar a la continuidad de $\exp$. En particular tenemos %#% $ #%

Por lo tanto el límite requerido es de $$\lim_{x\to a} \frac{\sin \left(\frac{\pi x}{2a}\right) \log\left(2 - \frac{x}{a}\right) }{\cos \left(\frac{\pi x}{2a}\right)} =\lim_{x\rightarrow a} \frac{\frac{\pi}{2a}\cos(\frac{\pi x}{2a})\log(2-\frac{x}{a})-\frac{1}{a}\left(2-\frac{x}{a}\right)^{-1}\sin(\frac{\pi x}{2a})}{-\frac{\pi}{2a}\sin(\frac{\pi x}{2a})} = \frac{-1/a}{-\pi/2a} = \frac{2}{\pi}.$

Answer 3

ps

Answer 4

$$A=\left(2-\frac{x}{a} \right)^{\tan( \frac{\pi x}{2a})}\implies \log(A)=\tan( \frac{\pi x}{2a})\log\left(2-\frac{x}{a} \right)$$ Now, using Taylor expansion around $x = un$ $$\log\left(2-\frac{x}{a} \right)=-\frac{x-a}{a}-\frac{(x-a)^2}{2 a^2}+O\left((x-a)^3\right)$$ Using Laurent expansion around $x = un %#% $ #% $ combinando el producto $$\tan( \frac{\pi x}{2a})=-\frac{2 a}{\pi (x-a)}+\frac{\pi (x-a)}{6 a}+O\left((x-a)^3\right)$$$\log(A)=\frac{2}{\pi }+\frac{x-a}{\pi a}+O\left((x-a)^2\right)$% $ $ Taylor again $que muestra el límite y cómo se aborda.

Answer 5

Set \lim_{x\to $y=x-a$, entonces $$ un} \left (2-\frac {x} {un} \right) ^ {\tan\left (\frac {\pi x} {2a} \right)} = \lim_ {y\to0} \left (1-\frac {y} {un} \right) ^ {-\cot\left (\frac {\pi y} {2a} \right)} = \\ = \lim_ {y\to0} \exp \left(-\frac{\log\left(1-\frac{y}{a}\right)} {\tan\left (\frac {\pi y} {2a} \right)} \right) = \lim_ {y\to un} \exp {\left (\frac {y} {un} \cdot\frac {2a} {\pi y} \right)} = e ^ {2/\pi} $$