Question

$\lim_{x \to 0} \frac{cos ax- cos bx}{x}$ a resposta é: 0...

Passo a passo please.

Answer 1

Sabe-se que

$cos (A) - cos (B)=-2sen\begin{pmatrix}\dfrac{A+B}{2}\end{pmatrix}sen\begin{pmatrix}\dfrac{A-B}{2}\end{pmatrix}$

também, que

$\displaystyle\lim_{x\to 0} \dfrac{sen(nx)}{x} = n$

Então, temos:

$\displaystyle\lim_{x \to 0} \frac{cos ax- cos bx}{x} = \displaystyle\lim_{x \to 0} \dfrac{ - 2sen\begin{pmatrix} \frac{ax + bx}{2} \end{pmatrix}sen\begin{pmatrix} \frac{ax - bx}{2} \end{pmatrix}}{x} \\ = - 2\displaystyle\lim_{x \to 0} \dfrac{ sen\begin{pmatrix} \frac{ax + bx}{2} \end{pmatrix}sen\begin{pmatrix} \frac{ax - bx}{2} \end{pmatrix}}{x} \\ = - 2 \cdot\displaystyle\lim_{x \to 0} \dfrac{ sen\begin{pmatrix} \frac{(a + b)x}{2}\end{pmatrix}}{x} \cdot\displaystyle\lim_{x \to 0} sen\begin{pmatrix} \dfrac{(a - b)x}{2} \end{pmatrix} \\ = - 2 \cdot \dfrac{(a + b)}{2} \cdot0 \\ = 0$