Question

the limit represents the derivative of some function f at some number a. state such an f and a.
\\( \lim _ { h \
ightarrow 0 } \frac { \cos ( \pi + h ) + 1 } { h } \\)
\\( f ( x ) = \cos ( x ), a = \pi / 3 \\)
\\( f ( x ) = \cos ( x ), a = \pi \\)
\\( f ( x ) = \cos ( x ), a = \pi / 6 \\)
\\( f ( x ) = \cos ( x ), a = \pi / 4 \\)
\\( f ( x ) = \cos ( x ), a = 0 \\)

Explanation:

Step1: Recall the definition of the derivative

The definition of the derivative of a function \(y = f(x)\) at \(x=a\) is \(f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}\).

Step2: Compare with the given limit

We are given \(\lim_{h
ightarrow0}\frac{\cos(\pi+h)+1}{h}\).
Since \(\cos(\pi)=- 1\), we can rewrite the given limit as \(\lim_{h
ightarrow0}\frac{\cos(\pi + h)-\cos(\pi)}{h}\).
Comparing with \(f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}\), we have \(f(x)=\cos(x)\) and \(a = \pi\).

Answer:

\(f(x)=\cos(x)\), \(a=\pi\) (corresponding to the second option \(f(x)=\cos(x),a = \pi\))