Evaluate the limit of $\frac{f(2+h)-f(2)}{h}$ as $h$ approaches $0$ for $f(x) = \frac{1}{x^2}$

Question

Evaluate the limit of $\frac{f(2+h)-f(2)}{h}$ as $h$ approaches $0$ for $f(x) = \frac{1}{x^2}$

Answer 1

$$\frac{f(2+h)-f(2)}{h}=\frac{\frac1 {(2+h)^2}-\frac1{2^2}}{h}={\frac1{4h+4h^2+h^3}-\frac1{4h}}=\frac{4h-4h-4h^2-h^3}{4h^2(4+4h+h^2)}=-\frac{4+h}{4(4+4h+h^2)}$$ Take the limit as $h\to0$, and you get $-\frac 4{16}=-\frac14$

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Another method is to notice that this is the definition of the derivative of $f$ at $2$: $$\frac{df}{dx}=\lim_{h\to0}\left(\frac{f(x+h)-f(x)}{h}\right)$$ The derivative of your function $f(x)=\frac1{x^2}$ is $$\frac{df}{dx}=-\frac{2}{x^3}$$ Evaluating this at $x=2$ gives the same answer as above.

Answer 2

One has $$\lim_{h\to 0} \frac{\frac{1}{(2+h)^2}-\frac{1}{2^2}}{h} = \lim_{h\to 0} \frac{-4h-h^2}{4h(h+2)^2} = \lim_{h\to 0}\frac{-4-h}{4(2+h)^2} = \frac{-1}{4}.$$

Answer 3

Unclear if this is allowed, but the limit is simply the value of the derivative $-\dfrac2{x^3}$ at $x=2$.