Question

(a) the derivative of a function f at x, denoted by f(x), is f(x)=\\(\lim_{h\to0}\frac{f(x + h)-f(x)}{h}\\) provided that the limit exists. use the definition of the derivative of f at x to find the derivative of the given function f(x) when x = 4. f(4)=\\(\lim_{h\to0}\frac{\square-\left\frac{1}{x - 1}\
ight}{h}\\)

Explanation:

Step1: Substitute \(x = 4\) into the derivative formula

The derivative formula is \(f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}\). When \(x = 4\), we have \(f^{\prime}(4)=\lim_{h
ightarrow0}\frac{f(4 + h)-f(4)}{h}\). Given \(f(x)=\frac{1}{x - 1}\), so \(f(4)=\frac{1}{4 - 1}=\frac{1}{3}\) and \(f(4 + h)=\frac{1}{(4 + h)-1}=\frac{1}{h + 3}\).

Step2: Simplify the fraction

$$
LATEXBLOCK0
$$

Cancel out the non - zero \(h\) (since \(h
ightarrow0\) but \(h
eq0\) in the limit process), we get \(\frac{-1}{3(h + 3)}\)

Step3: Evaluate the limit

\(\lim_{h
ightarrow0}\frac{-1}{3(h + 3)}\). Substitute \(h = 0\) into \(\frac{-1}{3(h + 3)}\), we have \(\lim_{h
ightarrow0}\frac{-1}{3(h + 3)}=-\frac{1}{9}\)

Answer:

\(-\frac{1}{9}\)