`lim_(x to infty) [x-sqrt(x^(2) +x)]`=

Options

A. `1//2`
B. 1
C. `-1//2`
D. 0

Answer

To solve the limit \( \lim_{x \to \infty} \left( x - \sqrt{x^2 + x} \right) \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ L = \lim_{x \to \infty} \left( x - \sqrt{x^2 + x} \right) \]

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

### Step 3: Simplify the numerator
The numerator simplifies to:
\[ x^2 - (x^2 + x) = -x \]
So now we have:
\[ L = \lim_{x \to \infty} \frac{-x}{x + \sqrt{x^2 + x}} \]

### Step 4: Factor out \(x\) in the denominator
Now, we factor \(x\) out of the denominator:
\[ L = \lim_{x \to \infty} \frac{-x}{x \left( 1 + \sqrt{1 + \frac{1}{x}} \right)} \]
This simplifies to:
\[ L = \lim_{x \to \infty} \frac{-1}{1 + \sqrt{1 + \frac{1}{x}}} \]

### Step 5: Evaluate the limit
As \(x\) approaches infinity, \(\frac{1}{x}\) approaches 0. Thus, we have:
\[ L = \frac{-1}{1 + \sqrt{1 + 0}} = \frac{-1}{1 + 1} = \frac{-1}{2} \]

### Final Answer
Therefore, the limit is:
\[ \boxed{-\frac{1}{2}} \]