Question

1. $lim_{h \to 0} \frac{ln(e + h) - 1}{h}$ is

(a) $f(e)$, where $f(x)=ln x$
(b) $f(e)$, where $f(x)=\frac{ln x}{x}$
(c) $f(1)$, where $f(x)=ln x$
(d) $f(1)$, where $f(x)=ln(x + e)$
(e) $f(0)$, where $f(x)=ln x$
calculator allowed

Explanation:

Step1: Recall the definition of the derivative

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

Step2: Analyze the given limit

We are given \(\lim_{h
ightarrow0}\frac{\ln(e + h)-1}{h}\). Since \(\ln(e)=1\), we can rewrite the limit as \(\lim_{h
ightarrow0}\frac{\ln(e + h)-\ln(e)}{h}\).

Step3: Compare with the derivative formula

If we let \(f(x)=\ln(x)\) and \(a = e\), then by the formula \(f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}\), we have \(f^{\prime}(e)=\lim_{h
ightarrow0}\frac{\ln(e + h)-\ln(e)}{h}\).

Answer:

A. \(f^{\prime}(e)\), where \(f(x)=\ln x\)