Question

lim┬(x→0)⁡〖(tg 3x)/(sen 4x)〗

Answer 1

Serão descritas abaixo três formas de resolver:

1ª forma) Inicialmente, manipulando a expressão dentro do limite (que será chamada de E):

$E=\dfrac{\tan(3x)}{\sin(4x)}=\dfrac{1}{\sin(4x)}\cdot\tan(2x+x)=\\\\=\dfrac{1}{\sin(4x)}\cdot\dfrac{\tan 2x+\tan x}{1-\tan x\tan 2x}=\\\\=\left(\dfrac{\tan 2x+\tan x}{\sin(4x)}\right)\cdot\dfrac{1}{1-\tan x\tan 2x}=\\\\=\left(\dfrac{\tan 2x}{\sin(4x)}+\dfrac{\tan x}{\sin(4x)}}\right)\cdot\dfrac{1}{1-\tan x\tan 2x}$

Usando que $\tan y=\dfrac{\sin y}{\cos y}$, que $\sin(2y)=2\sin(y)\cos(y)$ e que $\sin(4y)=2\sin(2y)\cos(2y)=2(2\sin(y)\cos(y))\cos(2y)$:

$\left(\dfrac{\frac{\sin(2x)}{\cos(2x)}}{2\sin(2x)\cos (2x)}+\dfrac{\frac{\sin(x)}{\cos(x)}}{2(2\sin (x)\cos (x))\cos (2x)}\right)\cdot\dfrac{1}{1-\tan x\tan 2x}=\\\\ =\left(\dfrac{1}{2\cos^2(2x)}+\dfrac{1}{4\cos^2(x)\cos (2x)}\right)\cdot\dfrac{1}{1-\tan x\tan 2x}$

Assim, podemos dizer:

$\lim_{x\to 0} \dfrac{\tan(3x)}{\sin(4x)}=\\\\=\lim_{x\to 0} \left(\dfrac{1}{2\cos^2(2x)}+\dfrac{1}{4\cos^2(x)\cos (2x)}\right)\cdot\dfrac{1}{1-\tan x\tan 2x}=\\\\=\left(\dfrac{1}{2\cos^2(2\cdot0)}+\dfrac{1}{4\cos^2(0)\cos (2\cdot0)}\right)\cdot\dfrac{1}{1-\tan (0)\tan (2\cdot0)}=\\\\=\left(\dfrac{1}{2\cos^2(0)}+\dfrac{1}{4\cos^2(0)\cos (0)}\right)\cdot\dfrac{1}{1-\tan (0)\tan (0)}=\\\\=\left(\dfrac{1}{2\cdot1}+\dfrac{1}{4\cdot1^2\cdot1}\right)\cdot\dfrac{1}{1-0\cdot0}=$

$=\left(\dfrac{1}{2}+\dfrac{1}{4}\right)\cdot\dfrac{1}{1}}=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{2}{4}+\dfrac{1}{4}=\dfrac{3}{4}\\\\\boxed{\lim_{x \to0} \dfrac{\tan(3x)}{\sin(4x)} =\dfrac{3}{4}}$

2ª forma) Vamos multiplicar a expressão por 12x em cima e embaixo, obtendo:

$\lim_{x \to 0} \dfrac{\tan(3x)}{\sin(4x)}= \lim_{x \to 0} \dfrac{\tan(3x)}{\sin(4x)}\cdot\dfrac{12x}{12x}= \\\\=\lim_{x \to 0} \dfrac{\tan(3x)}{12x}\cdot\dfrac{12x}{\sin(4x)}=\lim_{x \to 0} \dfrac{\tan(3x)}{4x\cdot3x}\cdot\dfrac{3x\cdot4x}{\sin(4x)}=\\\\=\lim_{x \to 0} \dfrac{1}{4x}\cdot\dfrac{\tan(3x)}{3x}\cdot3x\cdot\dfrac{4x}{\sin(4x)}=\lim_{x \to 0} \dfrac{3}{4}\cdot\dfrac{\tan(3x)}{3x}\cdot\dfrac{4x}{\sin(4x)}=$

$=\lim_{x \to 0} \dfrac{3}{4}\cdot\dfrac{1}{\cos(3x)}\cdot\left(\dfrac{\sin(3x)}{3x}\right)\cdot\left(\dfrac{\sin(4x)}{4x}\right)^{-1}=$

Usando o limite fundamental $\lim_{y\to 0}\dfrac{\sin y}{y}=1$, temos:

$\lim_{x \to 0} \dfrac{3}{4}\cdot\dfrac{1}{\cos(3x)}\cdot\left(\dfrac{\sin(3x)}{3x}\right)\cdot\left(\dfrac{\sin(4x)}{4x}\right)^{-1}=\dfrac{3}{4}\cdot\dfrac{1}{\cos (0)}\cdot1\cdot1=\\\\=\dfrac{3}{4}\cdot1=\dfrac{3}{4}\\\\\Longrightarrow\boxed{\lim_{x\to0}\dfrac{\tan(3x)}{\sin(4x)}=\dfrac{3}{4}}$

3ª forma) Esta é a mais simples, aplicaremos a regra de L'Hôspital:

$\lim_{x \to 0} \dfrac{\tan(3x)}{\sin(4x)}=^{L'H\^ospital}= \lim_{x \to 0} \dfrac{3\sec^2(3x)}{4\cos(4x)}=\lim_{x \to 0} \dfrac{3}{4}\cdot\dfrac{\sec^2(3x)}{\cos(4x)}\\\\=\lim_{x \to 0} \dfrac{3}{4}\cdot\dfrac{1}{\cos^2(3x)\cdot\cos(4x)}=\dfrac{3}{4}\cdot\dfrac{1}{\cos^2(3\cdot0)\cdot\cos(4\cdot0)}=\\\\=\dfrac{3}{4}\cdot\dfrac{1}{\cos^2(0)\cdot\cos(0)}=\dfrac{3}{4}\cdot\dfrac{1}{1^2\cdot1}=\dfrac{3}{4}\cdot\dfrac{1}{1\cdot1}=\dfrac{3}{4}\\\\ \boxed{\lim_{x\to0}\dfrac{\tan(3x)}{\sin(4x)}=\dfrac{3}{4}}$