Find lim h -> 0 (f(x + h) - f(x))/h, when f(x) = x/(sqrt(1 + x))

To find the limit

\[
\lim_{h \to 0} \frac{f(x + h) - f(x)}{h},
\]

we'll substitute the function \( f(x) = \frac{x}{\sqrt{1 + x}} \) into the expression and simplify. Let's work through the steps.

### Step 1: Express the difference quotient
We first write the difference quotient using \( f(x + h) \) and \( f(x) \):

\[
\frac{f(x + h) - f(x)}{h} = \frac{\frac{x + h}{\sqrt{1 + (x + h)}} - \frac{x}{\sqrt{1 + x}}}{h}.
\]

Now we need to simplify the numerator.

### Step 2: Combine the terms in the numerator
To combine the terms in the numerator, we need a common denominator. The common denominator between \( \sqrt{1 + (x + h)} \) and \( \sqrt{1 + x} \) is:

\[
\text{common denominator} = \sqrt{1 + (x + h)} \cdot \sqrt{1 + x}.
\]

So, the difference becomes:

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

### Step 3: Simplify the fraction
The difference quotient is now:

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

Now, let's take the limit as \( h \to 0 \).

### Step 4: Apply the limit \( h \to 0 \)
As \( h \to 0 \), we use the following approximations:
- \( \sqrt{1 + (x + h)} \to \sqrt{1 + x} \),
- \( (x + h) \to x \).

Thus, the difference quotient simplifies to the derivative of the function \( f(x) = \frac{x}{\sqrt{1 + x}} \).

### Step 5: Compute the derivative using the quotient rule
Alternatively, we can apply the quotient rule to compute the derivative of \( f(x) \) directly. The quotient rule states:

\[
\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2},
\]

where \( u = x \) and \( v = \sqrt{1 + x} \).

- \( u' = 1 \),
- \( v' = \frac{1}{2\sqrt{1 + x}} \).

Substitute these into the quotient rule:

\[
f'(x) = \frac{(1)\sqrt{1 + x} - (x)\frac{1}{2\sqrt{1 + x}}}{(\sqrt{1 + x})^2}.
\]

Simplify the expression:

\[
f'(x) = \frac{\sqrt{1 + x} - \frac{x}{2\sqrt{1 + x}}}{1 + x}.
\]

Combine the terms in the numerator:

\[
f'(x) = \frac{2(1 + x) - x}{2\sqrt{1 + x}(1 + x)} = \frac{2 + 2x - x}{2(1 + x)^{3/2}} = \frac{2 + x}{2(1 + x)^{3/2}}.
\]

### Final Answer:
The derivative, and hence the value of the limit, is:

\[
\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = f'(x) = \frac{2 + x}{2(1 + x)^{3/2}}.
\]

Would you like more details on this, or have any questions? Here are some related questions:

1. How do you apply the quotient rule to more complex functions?
2. What are some alternative ways to compute limits of difference quotients?
3. How does the chain rule apply in derivatives involving square roots?
4. Can you visualize the behavior of the derivative for different values of \(x\)?
5. How does the power rule differ from the quotient rule?

**Tip:** Practice recognizing when a function is suitable for applying the quotient rule versus the product rule to simplify your work.

Find the limit and derivative of the function f(x) = x/(sqrt(1 + x)) as h approaches 0. This problem demonstrates the use of the quotient rule and limit definition to calculate the derivative. A detailed step-by-step solution is provided, suitable for undergraduate calculus students.

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