`lim_(x to oo) (sqrt(x + 1) - sqrt(x))` equals

Options

A. `oo`
B. 0
C. `-1`
D. 1

Answer

To solve the limit \( \lim_{x \to \infty} (\sqrt{x + 1} - \sqrt{x}) \), we can follow these steps:

### Step 1: Identify the limit
Let \( L = \lim_{x \to \infty} (\sqrt{x + 1} - \sqrt{x}) \).

### Step 2: Rationalize the expression
To simplify the expression, we can multiply and divide by the conjugate \( \sqrt{x + 1} + \sqrt{x} \):

\[ L = \lim_{x \to \infty} \frac{(\sqrt{x + 1} - \sqrt{x})(\sqrt{x + 1} + \sqrt{x})}{\sqrt{x + 1} + \sqrt{x}} \]

### Step 3: Apply the difference of squares
Using the identity \( a^2 - b^2 = (a - b)(a + b) \), we can simplify the numerator:

\[ L = \lim_{x \to \infty} \frac{(x + 1) - x}{\sqrt{x + 1} + \sqrt{x}} \]

This simplifies to:

\[ L = \lim_{x \to \infty} \frac{1}{\sqrt{x + 1} + \sqrt{x}} \]

### Step 4: Simplify the denominator
As \( x \) approaches infinity, both \( \sqrt{x + 1} \) and \( \sqrt{x} \) approach \( \sqrt{x} \). Therefore, we can write:

\[ \sqrt{x + 1} \approx \sqrt{x} \quad \text{as } x \to \infty \]

Thus, the denominator becomes:

\[ \sqrt{x + 1} + \sqrt{x} \approx \sqrt{x} + \sqrt{x} = 2\sqrt{x} \]

### Step 5: Substitute back into the limit
Now we can substitute this back into our limit:

\[ L = \lim_{x \to \infty} \frac{1}{2\sqrt{x}} \]

### Step 6: Evaluate the limit
As \( x \) approaches infinity, \( \sqrt{x} \) also approaches infinity, which means:

\[ L = \frac{1}{2 \cdot \infty} = 0 \]

### Conclusion
Thus, we find that:

\[ \lim_{x \to \infty} (\sqrt{x + 1} - \sqrt{x}) = 0 \]

### Final Answer
The limit is \( 0 \).

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