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[tex]\displaystyle \text{cara I}\\\lim_{h\to0}\frac{\frac{1}{(2+h)^2}-\frac{1}{2^2}}{h}=\lim_{h\to0}\frac{\frac{1}{4+4h+h^2}-\frac14}{h}\\\lim_{h\to0}\frac{\frac{1}{(2+h)^2}-\frac{1}{2^2}}{h}=\lim_{h\to0}\frac{\frac{4-4-4h-h^2}{4(4+4h+h^2)}}{h}\\\lim_{h\to0}\frac{\frac{1}{(2+h)^2}-\frac{1}{2^2}}{h}=\lim_{h\to0}\frac{-4h-h^2}{4h(4+4h+h^2)}\\\lim_{h\to0}\frac{\frac{1}{(2+h)^2}-\frac{1}{2^2}}{h}=\lim_{h\to0}\frac{-4-h}{4(4+4h+h^2)}[/tex]
[tex]\displaystyle \lim_{h\to0}\frac{\frac{1}{(2+h)^2}-\frac{1}{2^2}}{h}=\frac{-4-0}{4(4+4(0)+0^2)}\\\lim_{h\to0}\frac{\frac{1}{(2+h)^2}-\frac{1}{2^2}}{h}=\frac{-4}{4(4)}\\\boxed{\boxed{\lim_{h\to0}\frac{\frac{1}{(2+h)^2}-\frac{1}{2^2}}{h}=-\frac{1}{4}}}[/tex]

[tex]\displaystyle \text{cara II}\\\lim_{h\to0}\frac{f(2+h)-f(2)}{h}=f'(2)\\\lim_{h\to0}\frac{f(2+h)-f(2)}{h}=\left\frac{d}{dx}\left(\frac1{x^2}\right)\right|_2\\\lim_{h\to0}\frac{f(2+h)-f(2)}{h}=\left-\frac{2}{x^3}\right|_2\\\lim_{h\to0}\frac{f(2+h)-f(2)}{h}=-\frac{2}{2^3}\\\lim_{h\to0}\frac{f(2+h)-f(2)}{h}=-\frac{2}{8}\\\boxed{\boxed{\lim_{h\to0}\frac{f(2+h)-f(2)}{h}=-\frac14}}[/tex]