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The given limit expression is:

\[
\lim_{{h \to 0}} \left( \frac{\frac{1}{{a+h}} - \frac{1}{a}}{h} \right),
\]
where \(a\) is a non-zero real constant.

### Step-by-step Solution:
This is a difference quotient and resembles the definition of the derivative. Specifically, it seems to represent the derivative of \(f(x) = \frac{1}{x}\) at \(x = a\).

#### Step 1: Use the difference of fractions formula
We begin by simplifying the numerator:

\[
\frac{1}{a+h} - \frac{1}{a} = \frac{a - (a+h)}{a(a+h)} = \frac{-h}{a(a+h)}.
\]

Thus, the expression inside the limit becomes:

\[
\frac{\frac{-h}{a(a+h)}}{h} = \frac{-h}{h \cdot a(a+h)}.
\]

#### Step 2: Simplify the expression
The \(h\) terms cancel out:

\[
\frac{-h}{h \cdot a(a+h)} = \frac{-1}{a(a+h)}.
\]

#### Step 3: Take the limit as \(h \to 0\)
As \(h\) approaches 0, the term \(a+h\) approaches \(a\). So the limit becomes:

\[
\lim_{{h \to 0}} \frac{-1}{a(a+h)} = \frac{-1}{a^2}.
\]

### Final Answer:
\[
\lim_{{h \to 0}} \left( \frac{\frac{1}{{a+h}} - \frac{1}{a}}{h} \right) = \frac{-1}{a^2}.
\]

Would you like further details or clarifications on any steps?

---

Here are 5 related questions to expand your understanding:

1. How is this limit connected to the concept of a derivative?
2. Can you derive the same result using the definition of the derivative?
3. What would the result be if \(a = 1\)?
4. How would this limit change if we had \( f(x) = \frac{1}{x^2} \)?
5. Why is it important that \(a \neq 0\) in this problem?

### Tip:
In problems involving limits and fractions, simplifying the numerator first often helps make the limit evaluation much easier.

Find the limit of (1/(a+h) - 1/a) / h as h approaches 0, where a is a non-zero real-valued constant.

This problem explores the limit of the expression (1/(a+h) - 1/a) / h as h tends to 0, where a is a non-zero constant. This limit demonstrates how limits are used to find the derivative of rational functions, particularly for f(x) = 1/x. Suitable for students in undergraduate calculus.

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