use-the-following-theorem-to-evaluate-limx0sin39x9xdisplaystyle-limx-to-0fracsin

Question

Use the following theorem to evaluate lim⁡x→0sin⁡39x9x\displaystyle \lim_{x 	o 0}{\frac{\sin{39x}}{9x}}:lim⁡x→0sin⁡xx=1\displaystyle \lim_{x 	o 0}{\frac{\sin{x}}{x}}=1

Accepted Answer

133\frac{13}{3}

Answer

−3-3

Answer

1313

Answer

583\frac{58}{3}

Use the following theorem to evaluate limx→0sin39x9x\(\displaystyle\) \(\lim\)_{x \(\to\) 0}{\(\frac{\sin{39x}\)}{9x}}:limx→0sinxx=1\(\displaystyle\) \(\lim\)_{x \(\to\) 0}{\(\frac{\sin{x}\)}{x}}=1

Use the following theorem to evaluate $\lim_{x \to 0} \frac{\sin 39 x}{9 x}$ :
$\lim_{x \to 0} \frac{\sin x}{x} = 1$