Question

Sa se calculeze limita:

$\lim_{x \to \ 0 } \frac{xe^{\frac{-1}{x} } }{tg^2x}\\ x\ \textgreater \ 0$

Answer 1

$l =\lim\limits_{x\searrow 0}\dfrac{xe^{-\frac{1}{x}}}{\mathrm{tg}^2 x}= \lim\limits_{x\searrow 0}\Bigg[\Big(\dfrac{x}{\mathrm{tg}\,x}\Big)^2\cdot \dfrac{e^{-\frac{1}{x}}}{x}\Bigg] =\\ \\ =1\cdot \lim\limits_{x\searrow 0}\dfrac{e^{-\frac{1}{x}}}{x} =\lim\limits_{x\searrow 0}\dfrac{e^{-\frac{1}{x}}}{x}$

$\text{Fac schimbarea de variabila:} \\\\ e^{-\frac{1}{x}} = t \Rightarrow -\dfrac{1}{x} = \ln t \Rightarrow \dfrac{1}{x} = -\ln t \\ \\ x\searrow 0 \Rightarrow e^{-\frac{1}{x}}\searrow 0 \Rightarrow t\searrow 0$

$l=\lim\limits_{t\searrow 0}(-t\ln t) = -\lim\limits_{t\searrow 0}\ln(t^t) = -\ln\Big(\lim\limits_{t\searrow 0}t^t\Big) = -\ln 1 =0$