Tem-se $\lim_{x\rightarrow p}\frac{g(x)-g(p)}{x-p}$limx→p​x−pg(x)−g(p)​, em que $g(x) = \frac{1}{x^2}$g(x)=x21​. Assim,

$\lim_{x\rightarrow p}\frac{\frac{1}{x^2}-\frac{1}{p^2}}{x-p}=\lim_{x\rightarrow p}\frac{\frac{p^2-x^2}{(xp)^2}}{x-p}=\lim_{x\rightarrow p}\bigg[ \frac{(p-x)(p+x)}{(xp)^2} \cdot \frac{1}{x-p}\bigg] = \lim_{x\rightarrow p}\bigg[\frac{-\cancel{(x-p)}(p+x)}{(xp)^2}\cdot \frac{1}{\cancel{x-p}}\bigg]=\lim_{x\rightarrow p}-\frac{p+x}{(xp)^2}=-\frac{2p}{(p^2)^2}=-\frac{2}{p^3}.$limx→p​x−px21​−p21​​=limx→p​x−p(xp)2p2−x2​​=limx→p​[(xp)2(p−x)(p+x)​⋅x−p1​]=limx→p​[(xp)2−(x−p)​(p+x)​⋅x−p​1​]=limx→p​−(xp)2p+x​=−(p2)22p​=−p32​.

Portanto,

$\lim_{x\rightarrow p}\frac{g(x)-g(p)}{x-p}$limx→p​x−pg(x)−g(p)​, em que $g(x)=\frac{1}{x^2}$g(x)=x21​ é igual a $-\frac{2}{p^3}.$−p32​.

lim x →p g(x) - g(p)/x - p = (1/x² - 1/p²) / x - p

(1/x² - 1/p²) / x - p → simplificando = - (x + p) / p²x²

lim x→p -(x + p) / p²x² = - (p + p) / p²p² = - 2p / p⁴ = -2/p³