Question

Salut, cum il aflu pe g(x) in problema 495...??

Answer 1

$\displaystyle Nu~ne~trebuie~forma~lui~g~pentru~rezolvarea~problemei. \\ \\ (Si~ca~fapt~divers,~daca~tot~ai~intrebat:~g~nu~poate~fi~exprimat~in \\ \\ functie~de~functii~elementare.) \\ \\ Tot~ce~trebuie~sa~stim~este~ca~g(f(x))=x~\forall~ x \in \mathbb{R}. \\ \\ Deci~voi~nota~x=\ln f(t).$

$\displaystyle Pentru~x=0~avem~f(t_0)=1,~deci~t_0+e^{t_0}=1,~cu~solutia~unica \\ \\ (caci~f~este~injectiva)~t_0=0. \\ \\ Deci~x \to 0 \Leftrightarrow t \to 0. \\ \\ \lim_{x\to 0} \frac{\sin x}{g(e^x)}= \lim_{t \to 0} \frac{\sin( \ln f(t))}{g(f(t))}= \lim_{t \to 0} \frac{\sin(\ln(t+e^t))}{t}= \\ \\ = \lim_{t \to 0} \frac{\sin(\ln(t+e^t))}{\ln(t+e^t)} \cdot \frac{\ln(t+e^t)}{t}. \\ \\ \lim_{t \to 0} \frac{\sin(\ln(t+e^t))}{\ln(t+e^t)}= \lim_{u \to 0} \frac{\sin u}{u}=1.$

$\displaystyle \lim_{t \to 0} \frac{\ln(t+e^t)}{t}= L'Hospital = \lim_{t \to 0} \frac{1+e^t}{t+e^t}=1. \\ \\ Deci~raspunsul~este~1.$