`"lt"_(x to infty) sqrt(x) (sqrt(x+c) - sqrt(x))`=

Options

A. `c//2`
B. `c//3`
C. `c//4`
D. none

Answer

To solve the limit \( L = \lim_{x \to \infty} \sqrt{x} \left( \sqrt{x+c} - \sqrt{x} \right) \), we can follow these steps:

### Step 1: Rewrite the Expression
We start with the expression:
\[ L = \lim_{x \to \infty} \sqrt{x} \left( \sqrt{x+c} - \sqrt{x} \right) \]

### Step 2: Rationalize the Expression
To simplify the expression, we can multiply and divide by the conjugate:
\[ L = \lim_{x \to \infty} \sqrt{x} \left( \sqrt{x+c} - \sqrt{x} \right) \cdot \frac{\sqrt{x+c} + \sqrt{x}}{\sqrt{x+c} + \sqrt{x}} \]
This gives us:
\[ L = \lim_{x \to \infty} \frac{\sqrt{x} \left( x+c - x \right)}{\sqrt{x+c} + \sqrt{x}} = \lim_{x \to \infty} \frac{\sqrt{x} \cdot c}{\sqrt{x+c} + \sqrt{x}} \]

### Step 3: Simplify the Denominator
Now, we simplify the denominator:
\[ \sqrt{x+c} + \sqrt{x} = \sqrt{x(1 + \frac{c}{x})} + \sqrt{x} = \sqrt{x} \left( \sqrt{1 + \frac{c}{x}} + 1 \right) \]

### Step 4: Substitute Back into the Limit
Substituting this back into our limit, we have:
\[ L = \lim_{x \to \infty} \frac{\sqrt{x} \cdot c}{\sqrt{x} \left( \sqrt{1 + \frac{c}{x}} + 1 \right)} = \lim_{x \to \infty} \frac{c}{\sqrt{1 + \frac{c}{x}} + 1} \]

### Step 5: Evaluate the Limit
As \( x \to \infty \), \( \frac{c}{x} \to 0 \), so:
\[ \sqrt{1 + \frac{c}{x}} \to \sqrt{1} = 1 \]
Thus, we have:
\[ L = \frac{c}{1 + 1} = \frac{c}{2} \]

### Final Answer
The limit is:
\[ L = \frac{c}{2} \]

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