To find the limit of the expression as $x$ approaches 0:

$\lim_{{x \to 0}} \frac{1}{x} \times \left(\frac{1}{h + x} - \frac{1}{h}\right)$

We can simplify and evaluate the expression step by step.

Step 1: Simplify the difference inside the parentheses

We have:

$\frac{1}{h + x} - \frac{1}{h}$

To subtract these fractions, find a common denominator:

$\frac{1}{h + x} - \frac{1}{h} = \frac{h - (h + x)}{h(h + x)} = \frac{-x}{h(h + x)}$

Step 2: Substitute this back into the original expression

Now substitute the simplified expression back into the limit:

$\lim_{{x \to 0}} \frac{1}{x} \times \left(\frac{-x}{h(h + x)}\right)$

Step 3: Simplify further

The $x$ terms cancel out:

$\lim_{{x \to 0}} \frac{-1}{h(h + x)}$

Step 4: Evaluate the limit as $x$ approaches 0

Now, as $x$ approaches 0, the expression simplifies to:

$\lim_{{x \to 0}} \frac{-1}{h(h + 0)} = \frac{-1}{h^2}$

Final Answer

The limit is:

$\lim_{{x \to 0}} \frac{1}{x} \times \left(\frac{1}{h + x} - \frac{1}{h}\right) = -\frac{1}{h^2}$

Would you like more details on any step or have any questions?

Here are 5 related questions for further exploration:

1. How does the limit change if the expression inside the parentheses is different?
2. What happens if we replace $h$ with a function of $x$?
3. How would the limit be affected if $x$ approaches a different value instead of 0?
4. Can you generalize this result for higher-order differences?
5. What is the significance of the negative sign in the result?

Tip: When dealing with limits that involve fractions, finding a common denominator can simplify the process considerably.