Question

$\lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}$

Explanation:

Step1: Simplify the numerator

First, we simplify the numerator \(\frac{1}{(x + h)^2}-\frac{1}{x^2}\). Find a common denominator, which is \(x^2(x + h)^2\). Then we have:
\[

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

\]

Step2: Substitute back into the original limit

The original limit is \(\lim_{h
ightarrow0}\frac{\frac{1}{(x + h)^2}-\frac{1}{x^2}}{h}\). Substitute the simplified numerator we got above:
\[
\lim_{h
ightarrow0}\frac{\frac{-h(2x + h)}{x^2(x + h)^2}}{h}
\]
We can cancel out the \(h\) terms (since \(h
eq0\) when taking the limit as \(h
ightarrow0\)):
\[
\lim_{h
ightarrow0}\frac{- (2x + h)}{x^2(x + h)^2}
\]

Step3: Evaluate the limit as \(h

ightarrow0\)
Now, we substitute \(h = 0\) into the expression \(\frac{- (2x + h)}{x^2(x + h)^2}\):
\[
\frac{- (2x+0)}{x^2(x + 0)^2}=\frac{-2x}{x^4}=-\frac{2}{x^3}
\]

Answer:

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