 Let f(x) = x/(x2 + 1) . Find the limit of the average rate of
change of f(x) on the interval [2, c] as c approaches 2

The problem asks for the limit of the average rate of change of the function \( f(x) = \frac{x}{x^2 + 1} \) on the interval \([2, c]\) as \( c \) approaches 2.

### Step 1: Write the formula for the average rate of change
The average rate of change of a function \( f(x) \) over an interval \([a, b]\) is given by:

\[
\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}
\]

For this problem, \( a = 2 \) and \( b = c \). Therefore, the average rate of change is:

\[
\text{Average Rate of Change} = \frac{f(c) - f(2)}{c - 2}
\]

### Step 2: Find \( f(2) \) and \( f(c) \)
First, calculate \( f(2) \):

\[
f(2) = \frac{2}{2^2 + 1} = \frac{2}{4 + 1} = \frac{2}{5}
\]

Next, express \( f(c) \):

\[
f(c) = \frac{c}{c^2 + 1}
\]

### Step 3: Substitute into the average rate of change formula
Substituting the values into the formula, we get:

\[
\text{Average Rate of Change} = \frac{\frac{c}{c^2 + 1} - \frac{2}{5}}{c - 2}
\]

### Step 4: Find the limit as \( c \) approaches 2
We need to compute the following limit:

\[
\lim_{c \to 2} \frac{\frac{c}{c^2 + 1} - \frac{2}{5}}{c - 2}
\]

This expression is in an indeterminate form \(\frac{0}{0}\), so we can apply L'Hôpital's Rule. However, before applying L'Hôpital's Rule, let's simplify the expression as much as possible by combining the fractions in the numerator:

\[
\frac{c}{c^2 + 1} - \frac{2}{5} = \frac{5c - 2(c^2 + 1)}{5(c^2 + 1)} = \frac{5c - 2c^2 - 2}{5(c^2 + 1)} = \frac{-2c^2 + 5c - 2}{5(c^2 + 1)}
\]

Thus, the expression becomes:

\[
\frac{\frac{-2c^2 + 5c - 2}{5(c^2 + 1)}}{c - 2} = \frac{-2c^2 + 5c - 2}{5(c^2 + 1)(c - 2)}
\]

Now, take the limit as \( c \) approaches 2:

\[
\lim_{c \to 2} \frac{-2c^2 + 5c - 2}{5(c^2 + 1)(c - 2)}
\]

### Step 5: Apply L'Hôpital's Rule
We differentiate the numerator and the denominator with respect to \( c \):

- The derivative of the numerator \( -2c^2 + 5c - 2 \) is \( -4c + 5 \).
- The derivative of the denominator \( 5(c^2 + 1)(c - 2) \) is obtained by using the product rule.

However, it is easier to evaluate the simplified expression directly:

\[
\lim_{c \to 2} \frac{-4c + 5}{5(2c + 2)} = \frac{-4(2) + 5}{5(2(2) + 2)} = \frac{-8 + 5}{5(4 + 2)} = \frac{-3}{30} = -\frac{1}{10}
\]

So, the limit is:

\[
\boxed{-\frac{1}{10}}
\]

This is the limit of the average rate of change of \( f(x) \) on the interval \([2, c]\) as \( c \) approaches 2.

Would you like more details or have any questions?

Here are 5 related questions you might consider:
1. How is the average rate of change different from the instantaneous rate of change?
2. What is the geometric interpretation of the average rate of change?
3. How can L'Hôpital's Rule be applied to more complex limits?
4. What is the significance of indeterminate forms in calculus?
5. Can you derive the limit using a different method, like the definition of the derivative?

**Tip:** Always try to simplify the expression before applying advanced methods like L'Hôpital's Rule to make the calculation easier.

Explore the limit of the average rate of change of the function f(x) = x/(x^2 + 1) on the interval [2, c] as c approaches 2. Includes step-by-step solution using L'Hôpital's Rule and mathematical derivations.

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