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Let's break down the problem:

Given the function \( f(x) = \frac{7}{x} \):

### Part (a):
You are asked to determine which of the provided limits corresponds to the slope of the tangent line to the graph of \( f(x) \) at \( x = 3 \).

The slope of the tangent line is given by the derivative of the function at that point, so we need to compute \( f'(x) \) using the definition of the derivative:

\[
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
\]

For \( f(x) = \frac{7}{x} \), we substitute this into the limit formula:

\[
f'(x) = \lim_{h \to 0} \frac{\frac{7}{x+h} - \frac{7}{x}}{h}
\]

Now, we need to evaluate this derivative at \( x = 3 \):

\[
f'(3) = \lim_{h \to 0} \frac{\frac{7}{3+h} - \frac{7}{3}}{h}
\]

This matches the expression given in option **(B)**:

\[
\lim_{h \to 0} \frac{\frac{7}{3+h} - \frac{7}{3}}{h}
\]

So, the correct answer for part (a) is **(B)**.

### Part (b):
Now, we need to evaluate the limit to find the slope of the tangent line and use it to form the equation of the tangent line at \( x = 3 \).

From part (a), we already know that the slope of the tangent line is given by \( f'(3) \), so we evaluate the limit:

\[
f'(3) = \lim_{h \to 0} \frac{\frac{7}{3+h} - \frac{7}{3}}{h}
\]

To solve this, first simplify the expression inside the limit. Combine the two fractions in the numerator:

\[
\frac{7}{3+h} - \frac{7}{3} = 7 \left( \frac{1}{3+h} - \frac{1}{3} \right)
\]

Now, combine the two terms into a single fraction:

\[
= 7 \left( \frac{3 - (3+h)}{(3+h)(3)} \right) = 7 \left( \frac{-h}{(3+h)(3)} \right) = \frac{-7h}{3(3+h)}
\]

Thus, the limit becomes:

\[
f'(3) = \lim_{h \to 0} \frac{\frac{-7h}{3(3+h)}}{h}
\]

Simplifying further:

\[
= \lim_{h \to 0} \frac{-7}{3(3+h)} = \frac{-7}{9}
\]

So, the slope of the tangent line is \( -\frac{7}{9} \).

Now, we need to find the equation of the tangent line. The equation of the tangent line at \( x = 3 \) is given by:

\[
y = mx + b
\]

where \( m = f'(3) = -\frac{7}{9} \), and to find \( b \), we use the point \( (3, f(3)) \). Since \( f(3) = \frac{7}{3} \), the point is \( (3, \frac{7}{3}) \).

Substitute into the tangent line equation:

\[
\frac{7}{3} = -\frac{7}{9}(3) + b
\]

Simplifying:

\[
\frac{7}{3} = -\frac{7}{3} + b
\]

\[
b = \frac{7}{3} + \frac{7}{3} = \frac{14}{3}
\]

Thus, the equation of the tangent line is:

\[
y = -\frac{7}{9}x + \frac{14}{3}
\]

Let me know if you would like more details or if you have any questions! Here are 5 related questions you can explore:

1. How would you find the derivative of a more complex rational function?
2. What is the geometric interpretation of a tangent line on a curve?
3. How does the derivative relate to the slope of a secant line?
4. Can you solve a similar problem using implicit differentiation?
5. How does the derivative behave as \( x \to \infty \) for rational functions?

**Tip:** Always double-check your algebra when simplifying fractions inside a limit!

Let f(x) = 7/x. (a) Which of the given limits is equal to the slope of the tangent line to the graph of f(x) at x = 3? (b) Evaluate the limit in (a) and then use the result to find the equation of the tangent line to the graph of f(x) at x = 3.

Learn how to find the slope of the tangent line for f(x) = 7/x at x = 3 using limits and derivatives. This problem involves calculating the derivative of a rational function and forming the equation of the tangent line. Suitable for calculus students in Grades 11-12 or college.

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