Answer
$$\lim_{x\to-1}\frac{x^{2/9}-1}{x+1}=-\frac{2}{9}$$
Work Step by Step
$$A=\lim_{x\to-1}\frac{x^{2/9}-1}{x+1}=\lim_{x\to-1}\frac{x^{2/9}-(-1)^{2/9}}{x-(-1)}$$
According to the definition of derivative, we know that $$\lim_{x\to z}\frac{f(x)-f(z)}{x-z}=f'(z)$$
So here if we take $f(x)=x^{2/9}$, we have $$f'(z)=\lim_{x\to z}\frac{x^{2/9}-z^{2/9}}{x-z}=\frac{d(z^{2/9})}{dz}=\frac{2}{9}z^{2/9-1}=\frac{2}{9}z^{-7/9}$$
Take $z=-1$, then $$f'(-1)=\lim_{x\to1}\frac{x^{2/9}-(-1)^{2/9}}{x-(-1)}=\frac{2}{9}\times(-1)^{-7/9}=\frac{2}{9}\times(-1)=-\frac{2}{9}$$ $$A=-\frac{2}{9}$$
