Question

Afla valoarea limitei​

Answer 1

Raspuns:

$\displaystyle\lim_{x\to0}\left(\frac{\sqrt{x^2-4x+9}}{x+3}\right)^{\dfrac{1}{x}}=e^{-5/9}$

$\displaystyle\lim_{x\to0}\left(\frac{\sqrt{x^2-4x+9}}{x+3}\right)^{\dfrac{1}{x}}\overset{not}{=}E$

$E=\displaystyle\lim_{x\to0}\left(\frac{\sqrt{x^2-4x+9}}{\sqrt{(x+3)^2}}\right)^{\dfrac{1}{x}}=\displaystyle\lim_{x\to0}\left(\sqrt{\frac{x^2-4x+9}{(x+3)^2}}\right)^{\dfrac{1}{x}}=\displaystyle\lim_{x\to0}\left[\left(\frac{x^2-4x+9}{(x+3)^2}\right)^{\dfrac{1}{2}}\right]^{\dfrac{1}{x}}$

$=\displaystyle\lim_{x\to0}\left(\frac{x^2-4x+9}{(x+3)^2}\right)^{\dfrac{1}{2x}}=\displaystyle\lim_{x\to0}\left(\frac{x^2-4x+9}{x^2+6x+9}\right)^{\dfrac{1}{2x}}=\displaystyle\lim_{x\to0}\left(1+\frac{x^2-4x+9}{x^2+6x+9}-1\right)^{\dfrac{1}{2x}}=$

$=\displaystyle\lim_{x\to0}\left(1+\frac{x^2-4x+9-x^2-6x-9}{x^2+6x+9}\right)^{\dfrac{1}{2x}}=\displaystyle\lim_{x\to0}\left(1+\frac{-10x}{x^2+6x+9}\right)^{\dfrac{1}{2x}}=$

$=\displaystyle\lim_{x\to0}\left(1+\frac{-10x}{x^2+6x+9}\right)^{\dfrac{x^2+6x+9}{-10x}\cdot\dfrac{-10x}{x^2+6x+9}\cdot\dfrac{1}{2x}}=$

$=\displaystyle\lim_{x\to0}\left[\left(1+\frac{-10x}{x^2+6x+9}\right)^{\dfrac{x^2+6x+9}{-10x}}\right]^{\dfrac{-10x}{x^2+6x+9}\cdot\dfrac{1}{2x}}$

$\text{Putem scrie} \quad \dfrac{-10x}{x^2+6x+9} \quad \text{ca} \quad f(x)=\dfrac{-10x}{x^2+6x+9} \\\\\displaystyle\lim_{x\to0}f(x)=\lim_{x\to0}\dfrac{-10x}{x^2+6x+9} =\dfrac{0}{9}=0\\\\\text{Folosim urmatoarea limita remarcabila:}\\\\\displaystyle\lim_{f(x)\to0}(1+f(x))^{1/f(x)}=e$

$E=\displaystyle\lim_{x\to0}\left[\left(1+f(x)\right)^{\dfrac{1}{f(x)}\right]^{\dfrac{-10x}{x^2+6x+9}\cdot\dfrac{1}{2x}}=$

$=e^{\lim_{x\to0}\frac{-10x}{x^2+6x+9}\cdot\frac{1}{2x}}=e^{\lim_{x\to0}\frac{-5}{x^2+6x+9}}=e^{-\frac59}$