Question

approximate the limit of the difference quotient, $lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$, using $h=pm0.1,pm0.01$. $f(x)=\frac{1}{x + 1},a = 2$

Explanation:

Step1: First find $f(a + h)$ and $f(a)$

Given $f(x)=\frac{1}{x + 1}$ and $a = 2$. Then $f(a)=f(2)=\frac{1}{2+1}=\frac{1}{3}$, and $f(a + h)=f(2 + h)=\frac{1}{(2 + h)+1}=\frac{1}{h + 3}$.

Step2: Calculate the difference - quotient

The difference - quotient $\frac{f(a + h)-f(a)}{h}=\frac{\frac{1}{h + 3}-\frac{1}{3}}{h}=\frac{\frac{3-(h + 3)}{3(h + 3)}}{h}=\frac{\frac{3 - h - 3}{3(h + 3)}}{h}=\frac{\frac{-h}{3(h + 3)}}{h}=-\frac{1}{3(h + 3)}$.

Step3: Approximate the limit for different values of $h$

When $h = 0.1$:
Substitute $h = 0.1$ into $-\frac{1}{3(h + 3)}$, we get $-\frac{1}{3(0.1+3)}=-\frac{1}{3\times3.1}=-\frac{1}{9.3}\approx - 0.1075$.
When $h=-0.1$:
Substitute $h=-0.1$ into $-\frac{1}{3(h + 3)}$, we get $-\frac{1}{3(-0.1 + 3)}=-\frac{1}{3\times2.9}=-\frac{1}{8.7}\approx - 0.1149$.
When $h = 0.01$:
Substitute $h = 0.01$ into $-\frac{1}{3(h + 3)}$, we get $-\frac{1}{3(0.01+3)}=-\frac{1}{3\times3.01}=-\frac{1}{9.03}\approx - 0.1107$.
When $h=-0.01$:
Substitute $h=-0.01$ into $-\frac{1}{3(h + 3)}$, we get $-\frac{1}{3(-0.01 + 3)}=-\frac{1}{3\times2.99}=-\frac{1}{8.97}\approx - 0.1115$.

Answer:

When $h = 0.1$, the value is approximately $-0.1075$; when $h=-0.1$, the value is approximately $-0.1149$; when $h = 0.01$, the value is approximately $-0.1107$; when $h=-0.01$, the value is approximately $-0.1115$.