limit of 1 over x multiply ny 1 over h plus x - 1 over h as x approaches to 0

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.

Learn how to evaluate the limit of 1/x times (1/(h + x) - 1/h) as x approaches 0 with detailed steps. This problem involves limits and fractional calculus, focusing on the definition of limits and fraction subtraction. Suitable for advanced high school students and beyond.

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