Respuesta :

[tex]f'(x_0)=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=\dfrac{x}{6};\ x_0=-2\\\\subtitute\\\\f'(-2)=\lim\limits_{x\to-2}\dfrac{\frac{x}{6}-\frac{-2}{6}}{x-(-2)}=\lim\limits_{x\to-2}\dfrac{\frac{x}{6}+\frac{2}{6}}{x+2}=\lim\limits_{x\to-2}\dfrac{\frac{x+2}{6}}{x+2}\\\\=\lim\limits_{x\to-2}\left(\dfrac{x+2}{6}\cdot\dfrac{1}{x+2}\right)=\lim\limits_{x\to-2}\dfrac{1}{6}=\dfrac{1}{6}[/tex]