`lim_(h to 0) (sqrt(x+h)-sqrt(x))/h` is:

Options

A. `1/sqrt(x)`
B. `sqrt(x)`
C. `2sqrt(x)`
D. `1/(2sqrt(x))`

Answer

To solve the limit \( \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \), we can follow these steps:

### Step 1: Identify the form of the limit
As \( h \) approaches 0, both the numerator and denominator approach 0, resulting in an indeterminate form \( \frac{0}{0} \).

**Hint:** Check if substituting \( h = 0 \) leads to an indeterminate form.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \( \lim_{h \to c} \frac{f(h)}{g(h)} \) is of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then:
\[ \lim_{h \to c} \frac{f(h)}{g(h)} = \lim_{h \to c} \frac{f'(h)}{g'(h)} \]
In our case, let \( f(h) = \sqrt{x+h} - \sqrt{x} \) and \( g(h) = h \).

**Hint:** Differentiate the numerator and denominator separately.

### Step 3: Differentiate the numerator and denominator
1. Differentiate \( f(h) = \sqrt{x+h} - \sqrt{x} \):
\[ f'(h) = \frac{1}{2\sqrt{x+h}} \quad \text{(using the chain rule)} \]
2. Differentiate \( g(h) = h \):
\[ g'(h) = 1 \]

### Step 4: Rewrite the limit using derivatives
Now, we can rewrite the limit using the derivatives:
\[ \lim_{h \to 0} \frac{f'(h)}{g'(h)} = \lim_{h \to 0} \frac{\frac{1}{2\sqrt{x+h}}}{1} = \lim_{h \to 0} \frac{1}{2\sqrt{x+h}} \]

**Hint:** Substitute \( h = 0 \) in the derivative limit.

### Step 5: Evaluate the limit
Now, substitute \( h = 0 \):
\[ \lim_{h \to 0} \frac{1}{2\sqrt{x+h}} = \frac{1}{2\sqrt{x+0}} = \frac{1}{2\sqrt{x}} \]

### Final Answer
Thus, the limit is:
\[ \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} = \frac{1}{2\sqrt{x}} \]

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