Evaluate the following limits :
` Lim_(x to oo) sqrt(x^(2) +x-1) - x`

Answer

To evaluate the limit \( \lim_{x \to \infty} \left( \sqrt{x^2 + x - 1} - x \right) \), we can follow these steps:

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

### Step 2: Rationalize the Expression
To simplify the expression, we multiply by the conjugate:
\[ \lim_{x \to \infty} \left( \sqrt{x^2 + x - 1} - x \right) \cdot \frac{\sqrt{x^2 + x - 1} + x}{\sqrt{x^2 + x - 1} + x} \]

### Step 3: Simplify the Numerator
Using the difference of squares:
\[ \lim_{x \to \infty} \frac{(\sqrt{x^2 + x - 1})^2 - x^2}{\sqrt{x^2 + x - 1} + x} \]
This simplifies to:
\[ \lim_{x \to \infty} \frac{x^2 + x - 1 - x^2}{\sqrt{x^2 + x - 1} + x} \]
which further simplifies to:
\[ \lim_{x \to \infty} \frac{x - 1}{\sqrt{x^2 + x - 1} + x} \]

### Step 4: Factor Out \( x \) in the Denominator
In the denominator, we factor out \( x \):
\[ \sqrt{x^2 + x - 1} = \sqrt{x^2(1 + \frac{1}{x} - \frac{1}{x^2})} = x\sqrt{1 + \frac{1}{x} - \frac{1}{x^2}} \]
Thus, the limit becomes:
\[ \lim_{x \to \infty} \frac{x - 1}{x(\sqrt{1 + \frac{1}{x} - \frac{1}{x^2}} + 1)} \]

### Step 5: Simplify the Limit
Now we can cancel \( x \) in the numerator and denominator:
\[ \lim_{x \to \infty} \frac{1 - \frac{1}{x}}{\sqrt{1 + \frac{1}{x} - \frac{1}{x^2}} + 1} \]

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

### Final Answer
Thus, the limit is:
\[ \lim_{x \to \infty} \left( \sqrt{x^2 + x - 1} - x \right) = \frac{1}{2} \]

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