Answer
$$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=-\frac{1}{4}$$
Work Step by Step
$$f(x)=\frac{1}{x}\hspace{1cm} x=-2$$
So $f(x+h)=\frac{1}{x+h}$
Therefore, $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}$$ $$\lim_{h\to0}\frac{f(x+h)f(x)}{h}=\lim_{h\to0}\frac{\frac{x-(x+h)}{x(x+h)}}{h}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{x-x-h}{xh(x+h)}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{-h}{xh(x+h)}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{-1}{x(x+h)}$$
- Substitute $x=-2$ here: $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{-1}{(-2)(-2+h)}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\frac{-1}{(-2)(-2+0)}=\frac{-1}{(-2)(-2)}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=-\frac{1}{4}$$
