Question

openstax.org/books/calculus - volume - 1/pages/2 - 3 - the - limit - lim_{h→0}\frac{\frac{1}{(a + h)^2}-\frac{1}{a^2}}{h}

Explanation:

Step1: Combine the fractions in the numerator

$$\begin{align*} \frac{\frac{1}{(a + h)^2}-\frac{1}{a^2}}{h}&=\frac{\frac{a^2-(a + h)^2}{a^2(a + h)^2}}{h}\\ \end{align*}$$

Use the formula \((x + y)^2=x^{2}+2xy + y^{2}\), so \((a + h)^2=a^{2}+2ah+h^{2}\). Then \(a^2-(a + h)^2=a^2-(a^{2}+2ah + h^{2})=-2ah - h^{2}\)
The expression becomes \(\frac{\frac{-2ah - h^{2}}{a^2(a + h)^2}}{h}\)

Step2: Simplify the complex - fraction

\(\frac{\frac{-2ah - h^{2}}{a^2(a + h)^2}}{h}=\frac{-2ah - h^{2}}{a^2(a + h)^2}\cdot\frac{1}{h}\)
Factor out \(h\) from the numerator: \(\frac{h(-2a - h)}{a^2(a + h)^2}\cdot\frac{1}{h}\)
Cancel out the common factor \(h\) (since \(h
eq0\) as we are taking the limit as \(h
ightarrow0\)), we get \(\frac{-2a - h}{a^2(a + h)^2}\)

Step3: Evaluate the limit

\(\lim_{h
ightarrow0}\frac{-2a - h}{a^2(a + h)^2}\)
Substitute \(h = 0\) into the expression \(\frac{-2a-0}{a^2(a + 0)^2}\)
\(\frac{-2a}{a^2\cdot a^2}=-\frac{2}{a^3}\)

Answer:

\(-\frac{2}{a^3}\)