@ Mathstud, #15 : it looks complicated, huh ?
Correct substitution, but what about this approximation : $\displaystyle \cos x \sim 1-\frac{x^2}2$ (Taylor series, that's what I think about when talking about approximations) when $\displaystyle x \sim 0$
$\displaystyle t=x-1$
--> $\displaystyle \lim_{t \to 0} \frac{\ln \sin(\tfrac \pi 2+\tfrac \pi 2 t)}{t^2}=\lim_{t \to 0} \frac{\ln \cos(\tfrac \pi 2 t)}{t^2}=\lim_{t \to 0} \frac{\ln(1-\tfrac 12 \cdot \tfrac{\pi^2 t^2}4)}{t^2} \sim -\frac{\tfrac{\pi^2t^2}{8}}{t^2}=-\frac{\pi^2}{8}$
@ Chris : Yes
Let
$\displaystyle t=x+\frac{\pi}{6}$
So we get
$\displaystyle \lim_{t\to{0}}\frac{t^2}{2\sin\left(x+\frac{\pi}{6 }\right)-1}$
Now note that
$\displaystyle \sin\left(x+\frac{\pi}{6}\right)$
$\displaystyle =\sin\left(x-\frac{\pi}{3}+\frac{\pi}{2}\right)$
$\displaystyle =\cos\left(x-\frac{\pi}{3}\right)$
$\displaystyle =\frac{\cos(x)}{2}+\frac{\sqrt{3}\sin(x)}{2}$
so
$\displaystyle 2\sin\left(x+\frac{\pi}{6}\right)=\cos(x)+\sqrt{3} \sin(x)$
So we have
$\displaystyle \lim_{t\to{0}}\frac{t^2}{\cos(t)+\sqrt{3}\sin(t)+1 }=0\quad\blacksquare$
Yeah there goes that, and the Geometric Arithmetic mean thing is so nice. but here goes nothing
$\displaystyle L=\lim_{x\to{0}}\left(\frac{a^x+b^x}{2}\right)^{\f rac{1}{x}}$
$\displaystyle \Rightarrow\ln\left(L\right)=\lim_{x\to{0}}\frac{\ ln\left(\frac{a^x+b^x}{2}\right)}{x}$
$\displaystyle =\lim_{x\to{0}}\frac{\ln\left(a^xb^x\cdot\frac{b^{-x}+a^{-x}}{2}\right)}{x}$
$\displaystyle =\lim_{x\to{0}}\frac{\ln(a^xb^x)}{x}+\lim_{x\to{0} }\frac{\ln\left(\frac{a^{-x}+b^{-x}}{2}\right)}{x}$
$\displaystyle \Rightarrow{L=\ln(ab)+\lim_{x\to{0}}\frac{\ln\left (\frac{a^{-x}+b^{-x}}{2}\right)}{x}}$
Now for the second limt
If we let $\displaystyle t=-x$
We get
$\displaystyle -\lim_{t\to{0}}\frac{\ln\left(\frac{a^t+b^t}{2}\rig ht)}{t}=-\ln(L)$
So
$\displaystyle \ln(L)=\ln(ab)-\ln(L)$
$\displaystyle \Rightarrow{2\ln(L)=\ln(ab)}$
$\displaystyle \Rightarrow{\ln(L)=\ln\left(\sqrt{ab}\right)}$
$\displaystyle \Rightarrow{L=\sqrt{ab}}\quad\blacksquare$
Hooray! That was cool!
[tex]
Hi !
Let $\displaystyle f(x)=x-a\ln x$ and $\displaystyle g(x)=\sqrt{2ax-x^2}$
$\displaystyle f'(x)= 1-\frac{a}{x}$
$\displaystyle f''(x)=\frac{a}{x^2}$
$\displaystyle g'(x)=\frac{x-a}{\sqrt{2ax-x^2}}$
$\displaystyle \begin{aligned}g''(x)&=\frac{\sqrt{2ax-x^2}-(x-a)\cdot \frac{x-a}{\sqrt{2ax-x^2}}}{2ax-x^2}\\ &=\frac{2ax-x^2-x^2+2ax-a^2}{\sqrt{2ax-x^2}(2ax-x^2)}\\ &=\frac{-2x^2+4ax-a^2}{\sqrt{2ax-x^2}(2ax-x^2)}\end{aligned}$
$\displaystyle \frac{a-x-a \ln(a)+a \ln(x)}{a-(2ax-x^2)^{\tfrac 12}}=\frac{f(a)-f(x)}{g(a)-g(x)}\underset{a}{\sim} \frac{(x-a) f'(a)+\frac{(x-a)^2}{2}f''(a)}{(x-a) g'(a)+\frac{(x^-a)^2}{2}g''(a)}=\frac{f''(a)}{g''(a)}$ since $\displaystyle f'(a)=g'(a)=0$. As $\displaystyle f''(a)=-\frac{1}{a} $ and $\displaystyle g''(a)=\frac{2a^2-a^2}{a\times a^2}=\frac{1}{a}$,
$\displaystyle \boxed{\lim_{x \to a} \quad \frac{a-x-a \ln(a)+a \ln(x)}{a-(2ax-x^2)^{\tfrac 12}}=-\frac{\frac{1}{a}}{\frac{1}{a}}=-1}$