Question

cine ma ajuta la limitele de la ex. 341/340

Answer 1

$340)\,\,\,\,\lim\limits_{x\to 0}\left(\dfrac{(1+x)^{\frac{1}{x}}}{e}\right)^{\frac{1}{x}}=\\ =\lim\limits_{x\to 0}\left(\left(1+\dfrac{(1+x)^{\frac{1}{x}}-e}{e}\right)^{\dfrac{e}{(1+x)^{\frac{1}{x}}-e}}\right)^{\displaystyle \dfrac{(1+x)^{\frac{1}{x}}-e}{e}\cdot \frac{1}{x}} =$

$= e^{\displaystyle \lim\limits_{x\to 0} \dfrac{(1+x)^{\frac{1}{x}}-e}{ex}} = e^{\displaystyle \lim\limits_{x\to 0} \dfrac{e^{\frac{1}{x}\ln(1+x)}-e}{ex}} = e^{\displaystyle \lim\limits_{x\to 0} \frac{e}{e}\cdot\dfrac{e^{\frac{1}{x}\ln(1+x)-1}-1}{x}} =$

$=e^{\displaystyle \lim\limits_{x\to 0} \dfrac{e^{\frac{1}{x}\ln(1+x)-1}-1}{\frac{1}{x}\ln(1+x)-1}\cdot \lim\limits_{x\to 0}\dfrac{\frac{1}{x}\ln(1+x)-1}{x}} =e^{\displaystyle 1\cdot \lim\limits_{x\to 0}\dfrac{\frac{1}{x}\ln(1+x)-1}{x}} =$

$= e^{\displaystyle 1\cdot \lim\limits_{x\to 0}\dfrac{\frac{1}{x}\ln(1+x)-1}{x}} = e^{\displaystyle \lim\limits_{x\to 0}\dfrac{\ln(1+x)-x}{x^2}} \overset{[L'H.]}{=}$

$\overset{[L'H.]}{=}e^{\displaystyle \lim\limits_{x\to 0}\dfrac{\frac{1}{1+x}-1}{2x}} \overset{[L'H.]}{=}e^{\displaystyle \lim\limits_{x\to 0}\dfrac{-\frac{1}{(1+x)^2}}{2}} = e^{-\frac{1}{2}} = \boxed{\dfrac{1}{\sqrt{e}}}$

$341)\,\,\,\,\lim\limits_{x\to 0}\left(\dfrac{\cos x}{\cos 2x}\right)^{\frac{1}{x^2}} =\\ = \lim\limits_{x\to 0}\left(\left(1+\dfrac{\cos x-\cos 2x}{\cos 2x}\right)^{\dfrac{\cos 2x}{\cos x-\cos 2x}}\right)^{\dfrac{\cos x-\cos 2x}{\cos 2x}\cdot \dfrac{1}{x^2}} =$

$=e^{\displaystyle \lim\limits_{x\to 0}\dfrac{\cos x-\cos 2x}{x^2\cos 2x}} \overset{[L'H.]}{=} e^{\displaystyle \lim\limits_{x\to 0}\dfrac{-\sin x+2\sin 2x}{2x\cos 2x -2x^2\sin 2x}} \overset{[L'H.]}{=}$

$\overset{[L'H.]}{=} e^{\displaystyle\lim\limits_{x\to 0}\dfrac{-\cos x+ 4\cos 2x}{2\cos 2x-4x\sin 2x -...}} = e^{\dfrac{-1+4}{2}} = e^{\dfrac{3}{2}} = \boxed{e\sqrt{e}}$

$\textbf{M-am folosit de limitele remarcabile:}$

$\lim\limits_{x\to a}\Big(1+u(x)\Big)^{\dfrac{1}{u(x)}} = e,\,\,\,\,\text{dac\u{a} }u(x) \to 0$

$\lim\limits_{x\to a}\dfrac{e^{u(x)}-1}{u(x)} = 1,\,\,\,\,\text{dac\u{a} }u(x) \to 0$