(f±g)'＝f'±g'となる例

$$\begin{array}{rl}{f}^{\prime }(a)\pm g^{\prime }(a)& =\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}\pm \underset{h\to 0}{lim}\frac{g(a+h)-g(a)}{h}\\ & =\underset{h\to 0}{lim}(\frac{f(a+h)-f(a)}{h}\pm \frac{g(a+h)-g(a)}{h})\\ & =\underset{h\to 0}{lim}\frac{(f(a+h)-f(a))\pm (g(a+h)-g(a))}{h}\\ & =\underset{h\to 0}{lim}\frac{(f(a+h)\pm g(a+h))-(f(a)\pm g(a))}{h}\\ & =\underset{h\to 0}{lim}\frac{(f\pm g)(a+h)-(f\pm g)(a)}{h}\\ & =(f\pm g{)}^{\prime }(a).\end{array}$$
（証明終）