Question

(URGENTE) Calcule o limite de sen(20x) dividido por sen(301x) quando x tende a 0. Ajuda ai

Respost 20/301

Answer 1

Explicação passo-a-passo:

$\sf \displaystyle\lim_{\sf x~\to~0}~\dfrac{\sf sen~(20x)}{\sf sen~(301x)}$

Multiplicando o numerador e o denominador por $\sf \dfrac{20x\cdot301}{20x\cdot301}$:

$\displaystyle\lim_{\sf x~\to~0}~\dfrac{\sf sen~(20x)}{\sf sen~(301x)}$

$=\displaystyle\lim_{\sf x\to~0}~\dfrac{\sf sen~(20x)}{\sf sen~(301x)}\cdot\dfrac{\sf 20x\cdot301}{\sf 20x\cdot301}$

$=\displaystyle\lim_{\sf x\to~0}~\dfrac{\sf sen~(20x)}{\sf 20x}\cdot\dfrac{\sf 301x}{\sf sen~(301x)}\cdot\dfrac{\sf 20}{\sf 301}$

$=\displaystyle\lim_{\sf x\to~0}~\dfrac{\sf sen~(20x)}{\sf 20x}\cdot\left[\dfrac{\sf sen~(301x)}{\sf 301x}\right]^{-1}\cdot\dfrac{\sf 20}{\sf 301}$

$=\displaystyle\lim_{\sf x\to~0}~\dfrac{\sf sen~(20x)}{\sf 20x}\cdot\lim_{\sf x\to~0}~\left[\dfrac{\sf sen~(301x)}{\sf 301x}\right]^{-1}\cdot\dfrac{\sf 20}{\sf 301}$

Pelo limite fundamental trigonométrico:

• $\displaystyle\lim_{\sf x\to~0}~\dfrac{\sf sen~(20x)}{\sf 20x}=\sf 1$

• $\displaystyle\lim_{\sf x\to~0}~\left[\dfrac{\sf sen~(301x)}{\sf 301x}\right]^{\sf -1}=\sf 1^{\sf -1}=\sf 1$

Logo:

$\displaystyle\lim_{\sf x\to~0}~\dfrac{\sf sen~(20x)}{\sf sen~(301x)}$

$=\underbrace{\displaystyle\lim_{\sf x\to~0}~\dfrac{\sf sen~(20x)}{\sf 20x}}_{=1}\cdot\underbrace{\displaystyle\lim_{\sf x\to~0}~\left[\dfrac{\sf sen~(301x)}{\sf 301x}\right]^{-1}}_{=1}\cdot\dfrac{20}{301}$

$\sf =1\cdot1\cdot\dfrac{20}{301}$

$\boxed{\displaystyle\lim_{\sf x\to~0}~\dfrac{\sf sen~(20x)}{\sf sen~(301x)}=\dfrac{\sf 20}{\sf 301}}$