Question

1. give the limit definition of derivative and use it to find $f(1)$ for $f(x) = \frac{4}{x + 1}$.

Explanation:

Step1: Recall limit definition of derivative

The limit definition of the derivative of a function \( f(x) \) at a point \( x = a \) is given by:

$$ f'(a)=\lim_{h ightarrow0}\frac{f(a + h)-f(a)}{h} $$

Here, we need to find \( f'(1) \), so \( a = 1 \). First, we find \( f(1+h) \) and \( f(1) \) for the function \( f(x)=\frac{4}{x + 1} \).

Step2: Compute \( f(1+h) \) and \( f(1) \)

For \( f(1+h) \): Substitute \( x=1 + h \) into \( f(x) \), we get \( f(1 + h)=\frac{4}{(1 + h)+1}=\frac{4}{h + 2} \)
For \( f(1) \): Substitute \( x = 1 \) into \( f(x) \), we get \( f(1)=\frac{4}{1+1}=\frac{4}{2} = 2 \)

Step3: Substitute into the derivative formula

Now, substitute \( f(1+h) \) and \( f(1) \) into the limit definition:

$$ f'(1)=\lim_{h ightarrow0}\frac{\frac{4}{h + 2}-2}{h} $$

Simplify the numerator:

$$ \frac{4}{h + 2}-2=\frac{4-2(h + 2)}{h + 2}=\frac{4-2h-4}{h + 2}=\frac{-2h}{h + 2} $$

So the expression for \( f'(1) \) becomes:

$$ f'(1)=\lim_{h ightarrow0}\frac{\frac{-2h}{h + 2}}{h} $$

Simplify the complex fraction:

$$ \frac{\frac{-2h}{h + 2}}{h}=\frac{-2h}{h + 2}\times\frac{1}{h}=\frac{-2}{h + 2}\quad(h eq0) $$

Step4: Evaluate the limit

Now, we evaluate the limit as \( h
ightarrow0 \):

$$ f'(1)=\lim_{h ightarrow0}\frac{-2}{h + 2} $$

Substitute \( h = 0 \) into \( \frac{-2}{h + 2} \) (since the function \( \frac{-2}{h + 2} \) is continuous at \( h = 0 \)):

$$ \lim_{h ightarrow0}\frac{-2}{h + 2}=\frac{-2}{0+2}=- 1 $$

Answer:

The limit definition of the derivative at \( x = a \) is \( f'(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h} \), and \( f'(1)=\boxed{-1} \)