Question

analiza matem 55 poli

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$f(x) = \dfrac{x}{e^x-a},\quad a>0\\ \\ f(x)\text{ nu are asimptote verticale }\Rightarrow \exists \text{ asimptota verticala daca}\\\\e^x-a = 0 \Rightarrow e^x = a \Rightarrow x = \ln a\\ \\ \lim\limits_{x\to\ln a} \dfrac{x}{e^x-a} = \lim\limits_{x\to\ln a} \dfrac{1}{e^x}= \dfrac{1}{\ln a} \Rightarrow \ln a = 0 \Rightarrow a = 1\\\\ \text{Deoarece trebuie ca limita sa nu aiba sens ca sa} \\ \text{nu existe asimptota verticala.}$

$\\ \\ \text{Asimptota orizontala:} \\ \\\lim\limits_{x\to\infty}\dfrac{x}{e^x-1} = 0 \\ \\ \lim\limits_{x\to -\infty}\dfrac{x}{e^x-1} = 0 = \infty\\ \\\Rightarrow y = 0\text{ asimptota orizontala spre }+\infty \\ \\ \text{Asimptote oblice:} \\ \\y = mx+n \\ \\m=\lim\limits_{x\to \infty}\dfrac{f(x)}{x} = 0,\quad (F)$

$m = \lim\limits_{x\to -\infty}\dfrac{f(x)}{x} = \lim\limits_{x\to -\infty}\dfrac{x}{x(e^x-1)} = 0 = -1\\ \\n = \lim\limits_{x\to-\infty}\Big(f(x) -mx\Big) =\lim\limits_{x\to -\infty}\Big(\dfrac{x}{e^x-1} +x\Big)= \\ \\ =\lim\limits_{x\to -\infty}\dfrac{x+x(e^x-1)}{e^x-1} = 0\\ \\\\e^x = t \Rightarrow t\to 0 \Rightarrow x = \ln t\\ \\=\lim\limits_{x\to 0}\dfrac{\ln t+\ln t(t-1)}{t-1}=\lim\limits_{x\to 0}\dfrac{\ln t+\ln{t^t}-\ln t}{t-1} = \\ \\ = \dfrac{0+\ln 1-0}{0-1} = 0$

$\Rightarrow y = -x\text{ asimptota oblica spre }-\infty$

a) corect