Evaluate the following limits :
`Lim_(x to oo) [ x - sqrt((x^(2)+x))] `

Answer

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

### Step 1: Write the limit expression
We start with the limit 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:
\[ L = \lim_{x \to \infty} \left( x - \sqrt{x^2 + x} \right) \cdot \frac{x + \sqrt{x^2 + x}}{x + \sqrt{x^2 + x}} \]
This gives us:
\[ L = \lim_{x \to \infty} \frac{\left( x - \sqrt{x^2 + x} \right) \left( x + \sqrt{x^2 + x} \right)}{x + \sqrt{x^2 + x}} \]

### Step 3: Simplify the numerator
Using the difference of squares in the numerator:
\[ L = \lim_{x \to \infty} \frac{x^2 - (x^2 + x)}{x + \sqrt{x^2 + x}} = \lim_{x \to \infty} \frac{-x}{x + \sqrt{x^2 + x}} \]

### Step 4: Simplify the denominator
Now, simplify the denominator:
\[ \sqrt{x^2 + x} = \sqrt{x^2(1 + \frac{1}{x})} = x\sqrt{1 + \frac{1}{x}} \]
Thus, we can rewrite the denominator:
\[ x + \sqrt{x^2 + x} = x + x\sqrt{1 + \frac{1}{x}} = x(1 + \sqrt{1 + \frac{1}{x}}) \]

### Step 5: Substitute back into the limit
Substituting back into our limit expression:
\[ L = \lim_{x \to \infty} \frac{-x}{x(1 + \sqrt{1 + \frac{1}{x}})} = \lim_{x \to \infty} \frac{-1}{1 + \sqrt{1 + \frac{1}{x}}} \]

### Step 6: Evaluate the limit
As \( x \to \infty \), \( \frac{1}{x} \to 0 \):
\[ L = \frac{-1}{1 + \sqrt{1 + 0}} = \frac{-1}{1 + 1} = \frac{-1}{2} \]

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