Question

calcular sem aplicar L'HOSPITAL: $\lim_{x \to \ \pi } \frac{tan(1+cos(x))}{cos(tang(x))-1}$

Answer 1

$L=\displaystyle\lim\limits_{x\to \pi}\;\dfrac{\tan(1+\cos x)}{\cos(\tan x)-1}\\ \\ \\ =\lim\limits_{x\to \pi}\;\tan(1+\cos x)\cdot \dfrac{1}{\cos(\tan x)-1}\\ \\ \\ =\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{\cos (1+\cos x)}\cdot \dfrac{1}{\cos(\tan x)-1}$

$\bullet\;\;$ Multiplicando e dividindo por $(1+\cos x),$ temos

$=\displaystyle\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{1+\cos x}\cdot \dfrac{1}{\cos(1+\cos x)}\cdot \dfrac{1+\cos x}{\cos(\tan x)-1}$

$\bullet\;\;$ Multiplicando e dividindo por $[\cos(\tan x)+1],$ temos

$=\displaystyle\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{1+\cos x}\cdot \dfrac{1}{\cos(1+\cos x)}\cdot \dfrac{(1+\cos x)\cdot [\cos(\tan x)+1]}{[\cos(\tan x)-1]\cdot [\cos(\tan x)+1]}\\ \\ \\ =\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{1+\cos x}\cdot \dfrac{1}{\cos(1+\cos x)}\cdot \dfrac{(1+\cos x)\cdot [\cos(\tan x)+1]}{\cos^{2}(\tan x)-1}\\ \\ \\ =\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{1+\cos x}\cdot \dfrac{1}{\cos(1+\cos x)}\cdot \dfrac{(1+\cos x)\cdot [\cos(\tan x)+1]}{-\sin^{2}(\tan x)}$

$\bullet\;\;$ Multiplicando e dividindo por $(1-\cos x),$ temos

$=\displaystyle\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{1+\cos x}\cdot \dfrac{\cos(\tan x)+1}{\cos(1+\cos x)}\cdot \dfrac{(1+\cos x)\cdot (1-\cos x)}{-\sin^{2}(\tan x)\cdot (1-\cos x)}\\ \\ \\ =\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{1+\cos x}\cdot \dfrac{\cos(\tan x)+1}{\cos(1+\cos x)}\cdot \dfrac{1-\cos^{2}x}{-\sin^{2}(\tan x)\cdot (1-\cos x)}\\ \\ \\ =\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{1+\cos x}\cdot \dfrac{\cos(\tan x)+1}{\cos(1+\cos x)}\cdot \dfrac{\sin^{2}x}{-\sin^{2}(\tan x)\cdot (1-\cos x)}\\ \\ \\ =\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{1+\cos x}\cdot \dfrac{\cos(\tan x)+1}{\cos(1+\cos x)}\cdot \dfrac{1}{\cos x-1}\cdot \dfrac{\sin^{2}x}{\sin^{2}(\tan x)}$

$\bullet\;\;$ Multiplicando e dividindo por $x^{2},$ temos

$=\displaystyle\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{1+\cos x}\cdot \dfrac{\cos(\tan x)+1}{\cos(1+\cos x)}\cdot \dfrac{1}{\cos x-1}\cdot \dfrac{\sin^{2}x}{x^{2}}\cdot \dfrac{x^{2}}{\sin^{2}(\tan x)}\\ \\ \\ =\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{1+\cos x}\cdot \dfrac{\cos(\tan x)+1}{\cos(1+\cos x)}\cdot \dfrac{1}{\cos x-1}\cdot \left(\dfrac{\sin x}{x} \right )^{2}\cdot \dfrac{x^{2}}{\sin^{2}(\tan x)}$


$\bullet\;\;$ Multiplicando e dividindo por $\tan^{2}x,$ temos

$=\displaystyle\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{1+\cos x}\cdot \dfrac{\cos(\tan x)+1}{\cos(1+\cos x)}\cdot \dfrac{1}{\cos x-1}\cdot \\ \\ \\ \cdot \left(\dfrac{\sin x}{x} \right )^{2}\cdot \dfrac{x^{2}}{\sin^{2}(\tan x)}\cdot \dfrac{\tan^{2}x}{\tan^{2}x}\\ \\ \\ \\ =\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{1+\cos x}\cdot \dfrac{\cos(\tan x)+1}{\cos(1+\cos x)}\cdot \dfrac{1}{\cos x-1}\cdot \\ \\ \\ \cdot \left(\dfrac{\sin x}{x} \right )^{2}\cdot \dfrac{\tan^{2}x}{\sin^{2}(\tan x)}\cdot \dfrac{x^{2}}{\tan^{2}x}\\ \\ \\ \\ =\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{1+\cos x}\cdot \dfrac{\cos(\tan x)+1}{\cos(1+\cos x)}\cdot \dfrac{1}{\cos x-1}\cdot \\ \\ \\ \cdot \left(\dfrac{\sin x}{x} \right )^{2}\cdot \dfrac{1}{\left(\frac{\sin(\tan x)}{\tan x} \right )^{2}}\cdot \dfrac{x^{2}}{\tan^{2}x}$

$=\displaystyle\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{1+\cos x}\cdot \dfrac{\cos(\tan x)+1}{\cos(1+\cos x)}\cdot \dfrac{1}{\cos x-1}\cdot \\ \\ \\ \cdot \left(\dfrac{\sin x}{x} \right )^{2}\cdot \dfrac{1}{\left(\frac{\sin(\tan x)}{\tan x} \right )^{2}}\cdot \dfrac{\cos^{2}x}{\sin^{2}x}\cdot x^{2}\\ \\ \\ \\ =\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{1+\cos x}\cdot \dfrac{\cos(\tan x)+1}{\cos(1+\cos x)}\cdot \dfrac{1}{\cos x-1}\cdot \\ \\ \\ \cdot \left(\dfrac{\sin x}{x} \right )^{2}\cdot \dfrac{1}{\left(\frac{\sin(\tan x)}{\tan x} \right )^{2}}\cdot \dfrac{x^{2}}{\sin^{2}x}\cdot \cos^{2} x\\ \\ \\ \\ =\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{1+\cos x}\cdot \dfrac{\cos(\tan x)+1}{\cos(1+\cos x)}\cdot \dfrac{1}{\cos x-1}\cdot \\ \\ \\ \cdot \dfrac{1}{\left(\frac{\sin(\tan x)}{\tan x} \right )^{2}}\cdot \cos^{2} x$


Agora não tem mais nenhuma indeterminação:

$L=\lim\limits_{x\to \pi}\;\dfrac{\sin(1+\cos x)}{1+\cos x}\cdot\lim\limits_{x\to \pi}\; \dfrac{\cos(\tan x)+1}{\cos(1+\cos x)}\cdot \lim\limits_{x\to \pi}\;\dfrac{1}{\cos x-1}\cdot \\ \\ \\ \cdot \lim\limits_{x\to \pi}\;\dfrac{1}{\left(\frac{\sin(\tan x)}{\tan x} \right )^{2}}\cdot \lim\limits_{x\to \pi}\;\cos^{2} x\\ \\ \\ \\ =1\cdot \dfrac{2}{1}\cdot \dfrac{1}{(-2)}\cdot 1\cdot 1\\ \\ \\ =-1$