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.