Question

a. $\lim_{h\to0}\frac{(1 + h)^{-2}-1}{h}$

Explanation:

Step1: Recall the definition of the derivative

The derivative of a function \(y = f(x)\) at \(x = a\) is given by \(f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}\).
Let \(f(x)=x^{-2}=\frac{1}{x^{2}}\), and \(a = 1\). Then \(f(a+h)=(1 + h)^{-2}\) and \(f(a)=1^{-2}=1\).

Step2: Simplify the expression

First, expand \((1 + h)^{-2}=\frac{1}{(1 + h)^{2}}=\frac{1}{1 + 2h+h^{2}}\).

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Step3: Evaluate the limit

Now, find \(\lim_{h
ightarrow0}\frac{-2 - h}{(1 + h)^{2}}\).
Substitute \(h = 0\) into \(\frac{-2 - h}{(1 + h)^{2}}\), we get \(\frac{-2-0}{(1+0)^{2}}=- 2\).

Answer:

\(-2\)