Question

Ajuda !!
$lim \: \frac{x \: + \sin(x) }{{x}^{2} \: - \sin(x) } \\ x - > 0$
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Answer 1

Olá!

$\displaystyle\lim_{x\to0}\dfrac{x+\sin{x}}{x^2-\sin{x}}$

$=\displaystyle\lim_{x\to0}\dfrac{x\left(1+\dfrac{\sin{x}}{x}\right)}{x\left(x-\dfrac{\sin{x}}{x}\right)}$

$=\displaystyle\lim_{x\to0}\dfrac{1+\dfrac{\sin{x}}{x}}{x-\dfrac{\sin{x}}{x}}$

$=\dfrac{\displaystyle\lim_{x\to0}1+\displaystyle\lim_{x\to0}\dfrac{\sin{x}}{x}}{\displaystyle\lim_{x\to0}x-\displaystyle\lim_{x\to0}\dfrac{\sin{x}}{x}}$

$=\dfrac{1+1}{0-1}$

$=-2$

$\boxed{\displaystyle\lim_{x\to0}\dfrac{x+\sin{x}}{x^2-\sin{x}}=-2}$

Qualquer dúvida, comente! Bons estudos!