Question

problema 380 din culegerea de admitere la politehnica cluj
asimptota oblica

Va rog sa postati cu rezolvarea completa!
Cum calculez m=limita f(x)/x ?
imi da 1/infinit•0 si nu stiu sa rezolv mai departe

Answer 1

$\displaystyle f(x) = \frac{1}{1-e^{\frac{1}{x}}}, \,\,x\in (0,+\infty)$

$y = mx+n\,\,\,\text{(asimptot\u{a} oblic\u{a})}$

$\displaystyle m = \lim\limits_{x\to +\infty}\frac{f(x)}{x} = \lim\limits_{x\to +\infty}\frac{\frac{1}{1-e^{\frac{1}{x}}}}{x} = \lim\limits_{x\to +\infty}\left(\frac{1}{1-e^{\frac{1}{x}}}\cdot \frac{1}{x}\right) =$

$\displaystyle =\lim\limits_{x\to +\infty}\frac{\frac{1}{x}}{1-e^{\frac{1}{x}}}$

Fac schimbarea de variabilă:

$\frac{1}{x} = t \Rightarrow t\to \frac{1}{+\infty} \Rightarrow t\to 0_+$

$\displaystyle m = \underset{t>0}{\lim\limits_{t\to 0}}\,\frac{t}{1-e^t} \overset{\frac{0}{0}(L'H)}{=} \underset{t>0}{\lim\limits_{t\to 0}}\,\frac{t'}{(1-e^t)'} = \underset{t>0}{\lim\limits_{t\to 0}}\,\frac{1}{-e^t} = \frac{1}{-e^{0}} = \frac{1}{-1} = -1$

$\displaystyle n = \lim\limits_{x\to +\infty}\left(f(x)-mx\right) = \lim\limits_{x\to +\infty}\left(\frac{1}{1-e^{\frac{1}{x}}}-(-1)\cdot x\right) =$

$\displaystyle =\lim\limits_{x\to +\infty}\left(\frac{1}{1-e^{\frac{1}{x}}}+x\right) =\lim\limits_{x\to +\infty}\left(\frac{1}{1-e^{\frac{1}{x}}}+\frac{1}{\frac{1}{x}}\right) =$

Fac schimbarea de variabilă:

$\frac{1}{x} = t \Rightarrow t\to \frac{1}{+\infty} \Rightarrow t\to 0_+$

$\displaystyle =\underset{t>0}{\lim\limits_{t\to 0}}\left(\frac{1}{1-e^{t}}+\frac{1}{t}\right) = \underset{t>0}{\lim\limits_{t\to 0}}\,\frac{t+1-e^t}{t(1-e^{t})} \overset{\frac{0}{0}(L'H)}{=} \underset{t>0}{\lim\limits_{t\to 0}}\,\frac{1-e^t}{1-e^{t}-te^t} \overset{\frac{0}{0}(L'H)}{=}$

$\displaystyle \overset{\frac{0}{0}(L'H)}{=} \underset{t>0}{\lim\limits_{t\to 0}}\,\frac{-e^t}{-e^t-e^t-te^t} = \frac{-e^0}{-e^0-e^0-0\cdot e^0} = \frac{-1}{-1-1} = \frac{1}{2}$

$\Rightarrow \boxed{y = -x+\frac{1}{2}}$