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

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

We can simplify and evaluate the expression step by step.

Step 1: Simplify the expression inside the parentheses

We have:

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

To simplify this, multiply and divide by the conjugate of the numerator:

$\frac{1}{\sqrt{1+h}} - 1 = \frac{\left(\frac{1}{\sqrt{1+h}} - 1\right) \times \left(\frac{1}{\sqrt{1+h}} + 1\right)}{\frac{1}{\sqrt{1+h}} + 1}$

This simplifies to:

$\frac{1 - (\sqrt{1+h})}{\sqrt{1+h} \cdot \left(\frac{1}{\sqrt{1+h}} + 1\right)} = \frac{\sqrt{1+h} - 1}{h}$

Step 2: Substitute this back into the original expression

Now substitute the simplified expression back into the limit:

$\lim_{{h \to 0}} \frac{1}{h} \times \frac{\sqrt{1+h} - 1}{h}$

This becomes:

$\lim_{{h \to 0}} \frac{\sqrt{1+h} - 1}{h^2}$

Step 3: Simplify further using the derivative

Notice that the expression $\sqrt{1+h} - 1$ resembles the difference in the definition of a derivative. Specifically:

$\sqrt{1+h} \approx 1 + \frac{h}{2} \text{ for small } h$

So,

$\sqrt{1+h} - 1 \approx \frac{h}{2}$

Substituting back:

$\lim_{{h \to 0}} \frac{\frac{h}{2}}{h^2} = \lim_{{h \to 0}} \frac{1}{2h} = \infty$

This suggests that the original interpretation should include another step, specifically considering the first order approximation as:

] ### Final Answer The correct interpretation would be: \[ \lim_{{h \to 0}} \frac{1}{h} \times \frac{1}{\sqrt{1+h}} = \text{not defined}$$ If this is the case. Would you like more details on any step or have any questions? Here are 5 related questions for further exploration: 1. What happens if we expand using the Taylor series? 2. How does this process apply to higher order terms? 3. How does this limit change if there is an additional constant term? 4. How would the approach differ if $$h$$ approaches a negative value? 5. Can this limit be derived geometrically? **Tip:** Limits involving square roots often benefit from multiplying by the conjugate, which can simplify the expression and eliminate the square root.