To solve this problem, we'll go step by step through both parts:

Part (a) - Finding the derivative of $f(x) = \sqrt{x}$ using the limit definition of the derivative

The limit definition of the derivative is given by:

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

For $f(x) = \sqrt{x}$, we substitute this into the formula:

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

To simplify this expression, we multiply both the numerator and denominator by the conjugate of the numerator, which is $\sqrt{x+h} + \sqrt{x}$:

$f'(x) = \lim_{h \to 0} \frac{\left(\sqrt{x+h} - \sqrt{x}\right)\left(\sqrt{x+h} + \sqrt{x}\right)}{h \left(\sqrt{x+h} + \sqrt{x}\right)}$

This simplifies the numerator using the difference of squares:

$f'(x) = \lim_{h \to 0} \frac{(x+h) - x}{h \left(\sqrt{x+h} + \sqrt{x}\right)}$

$f'(x) = \lim_{h \to 0} \frac{h}{h \left(\sqrt{x+h} + \sqrt{x}\right)}$

The $h$'s cancel out:

$f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$

Now, taking the limit as $h \to 0$, we get:

$f'(x) = \frac{1}{2\sqrt{x}}$

So, the derivative of $f(x) = \sqrt{x}$ is:

$f'(x) = \frac{1}{2\sqrt{x}}$

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Part (b) - Finding the equation of the tangent line at $x = 9$

To find the equation of the tangent line, we need two things:

1. The slope of the tangent line at $x = 9$, which is given by $f'(9)$.
2. The point on the curve at $x = 9$, which is $f(9)$.

First, find the slope of the tangent line: $f'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{2 \times 3} = \frac{1}{6}$

Next, find the point on the curve at $x = 9$: $f(9) = \sqrt{9} = 3$

So the point is $(9, 3)$, and the slope of the tangent line is $\frac{1}{6}$.

Now, using the point-slope form of the equation of a line:

$y - y_1 = m(x - x_1)$

where $m = \frac{1}{6}$, $x_1 = 9$, and $y_1 = 3$. Substituting these values:

$y - 3 = \frac{1}{6}(x - 9)$

Simplifying:

$y - 3 = \frac{1}{6}x - \frac{9}{6}$

$y = \frac{1}{6}x - \frac{3}{2} + 3$

$y = \frac{1}{6}x + \frac{3}{2}$

Thus, the equation of the tangent line at $x = 9$ is:

$y = \frac{1}{6}x + \frac{3}{2}$

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Would you like a deeper explanation for any of the steps?

Here are 5 related questions to consider:

1. What is the geometric interpretation of the derivative at $x = 9$?
2. How would the derivative change if $f(x) = x^{1/3}$?
3. Can you use the power rule to differentiate $f(x) = \sqrt{x}$ and compare it with the result from the limit definition?
4. What is the second derivative of $f(x) = \sqrt{x}$, and what does it tell you about the concavity of the function?
5. How can you find the normal line at $x = 9$?

Tip: When finding a derivative using the limit definition, multiplying by the conjugate often helps simplify square roots.