Question

Calcule o limite abaixo sem utilizar L'Hôpital



$\mathsf{\displaystyle\lim_{x\to0} \dfrac{x - sen~x}{tg~x-x}}$

Answer 1

Podemos calcular o limite escrevendo as funções trigonométricas na forma de Série de Taylor no ponto para o qual x está tendendo:

Série de Taylor de $f(x)$ em $x=a$:

$\displaystyle f(x)=\sum_{k=0}^{\infty}\dfrac{f^{(k)}(a)}{k!}(x-a)^k$

Então, em $x=0$:

$\displaystyle \sin(x)=\sum_{k=0}^{\infty}\dfrac{\sin^{(k)}(0)}{k!}x^k\\\\ \sin(x)=\dfrac{\sin(0)}{0!}\cdot x^0+\dfrac{\cos(0)}{1!}\cdot x^1-\dfrac{\sin(0)}{2!}\cdot x^2-\dfrac{\cos(0)}{3!}\cdot x^3+o(x^3)\\\\ \sin(x)=x-\dfrac{x^3}{3!}+o(x^3)$

$\displaystyle \tan(x)=\sum_{k=0}^{\infty}\dfrac{\tan^{(k)}(0)}{k!}x^k\\\\ \tan(x)=\dfrac{\tan(0)}{0!}\cdot x^0+\dfrac{\sec^2(0)}{1!}\cdot x^1+\dfrac{2\sec(0)\cdot\tan(0)}{2!}\cdot x^2+\\+\dfrac{2[(\sec(0)\cdot\tan(0)\cdot\tan(0))+(\sec(0)\cdot\sec^2(0))]}{3!}\cdot x^3+o(x^3)\\\\ \tan(x)=x+\dfrac{x^3}{3}+o(x^3)$

Substituindo no limite dado:

$L=\lim_{x\to0}\dfrac{x-\sin(x)}{\tan(x)-x}\\\\ L=\lim_{x\to0}\dfrac{x-\left[x-\dfrac{x^3}{3!}+o(x^3)\right]}{\left[x+\dfrac{x^3}{3}+o(x^3)\right]-x}\\\\ L=\lim_{x\to0}\dfrac{\dfrac{x^3}{3!}-o(x^3)}{\dfrac{x^3}{3}+o(x^3)}$

Multiplicando em cima e embaixo por 3!:

$L=\lim_{x\to0}\dfrac{\dfrac{x^3}{3!}-o(x^3)}{\dfrac{x^3}{3}+o(x^3)}\\\\ L=\lim_{x\to0}\dfrac{\dfrac{x^3}{3!}\cdot3!-o(x^3)\cdot3!}{\dfrac{x^3}{3}\cdot3!+o(x^3)\cdot3!}\\\\ L=\lim_{x\to0}\dfrac{x^3-o(x^3)}{2x^3+o(x^3)}$

Multiplicando em cima e embaixo por x³:

$L=\lim_{x\to0}\dfrac{x^3-o(x^3)}{2x^3+o(x^3)}\\\\ L=\lim_{x\to0}\dfrac{\dfrac{x^3}{x^3}-\dfrac{o(x^3)}{x^3}}{\dfrac{2x^3}{x^3}+\dfrac{o(x^3)}{x^3}}\\\\ L=\lim_{x\to0}\dfrac{1-\dfrac{o(x^3)}{x^3}}{2+\dfrac{o(x^3)}{x^3}}\\\\ L=\dfrac{1}{2}\\\\ \boxed{\lim_{x\to0}\dfrac{x-\sin(x)}{\tan(x)-x}=\dfrac{1}{2}}$

Answer 2

$\sf \displaystyle \large{ \lim _{x\to \:0}\left(\frac{x-sen \left(x\right)}{tg \left(x\right)-x}\right)}\\\\\\=\lim _{x\to \:0}\left(\frac{1-\cos \left(x\right)}{\tan ^2\left(x\right)}\right)\\\\\\{Utilize\:a\:Identidade\:B\acute{a}sica\:da\:Trigonometria}:\quad \frac{1}{tan \left(x\right)}=cot \left(x\right)\\\\\\=\lim _{x\to \:0}\left(\cot ^2\left(x\right)\left(1-\cos \left(x\right)\right)\right)$

$\sf \displaystyle =\lim _{x\to \:0}\left(\cot ^2\left(x\right)\frac{\sin ^2\left(x\right)}{1+\cos \left(x\right)}\right)\\\\\\=\lim _{x\to \:0}\left(\frac{\cos ^2\left(x\right)}{1+\cos \left(x\right)}\right)\\\\\\=\frac{\cos ^2\left(0\right)}{1+\cos \left(0\right)}\\\\\\\to \boxed{\sf \frac{1}{2} }$