Limites trigonometricos utilizando limite fundamental

[Aprendiz007]

Como posso resolver estes limites com o conceito de limite fundamental?

Anexos

limite.png (1.52 KiB) Visualizado 1216 vezes

[danjr5]

Resolva o limite aplicando o conceito de limite fundamental: \(\lim_{x \to 0} \frac{x - \tan x}{x + \tan x}\)

Olá!

\(\mathsf{\lim_{x \to 0} \frac{x - \tan x}{x + \tan x} =}\)

\(\mathsf{\lim_{x \to 0} \frac{x - \frac{\sin x}{\cos x}}{x + \frac{\sin x}{\cos x}} =}\)

\(\mathsf{\lim_{x \to 0} \frac{\cos x \cdot x - \sin x}{\cos x} \div \frac{\cos x \cdot x + \sin x}{\cos x} =}\)

\(\mathsf{\lim_{x \to 0} \frac{\cos x \cdot x - \sin x}{\cos x} \cdot \frac{\cos x}{\cos x \cdot x + \sin x} =}\)

\(\mathsf{\lim_{x \to 0} \frac{\cos x \cdot x - \sin x}{\cos x \cdot x + \sin x} =}\)

Dividindo o numerador e o denominador por x, teremos:

\(\mathsf{\lim_{x \to 0} \frac{\frac{\cos x \cdot x - \sin x}{x}}{\frac{\cos x \cdot x + \sin x}{x}} =}\)

\(\mathsf{\lim_{x \to 0} \frac{x \cdot \left ( \cos x - \frac{\sin x}{x} \right )}{x \cdot \left ( \cos x + \frac{\sin x}{x} \right )} =}\)

\(\mathsf{\lim_{x \to 0} \frac{\cos x}{\cos x + \frac{\sin x}{x}} - \lim_{x \to 0} \frac{\frac{\sin x}{x}}{\cos x + \frac{\sin x}{x}} =}\)

Mas, sabemos que \(\mathsf{\lim_{x \to 0} \frac{\sin x}{x} = 1}\). Então,

\(\mathsf{\lim_{x \to 0} \frac{\cos x}{\cos x + \frac{\sin x}{x}} - \lim_{x \to 0} \frac{\frac{\sin x}{x}}{\cos x + \frac{\sin x}{x}} =}\)

\(\mathsf{\frac{1}{1 + 1} - \frac{1}{1 + 1} =}\)

\(\mathsf{\frac{1}{2} - \frac{1}{2} =}\)

\(\fbox{\mathsf{0}}\)