`lim_(xrarroo) (sqrt(x^2+2x-1)-x)=`

Options

A. `oo`
B. `1//2`
C. `4`
D. `1`

Answer

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

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

### Step 2: Multiply and Divide by the Conjugate
To simplify the expression, we multiply and divide by the conjugate:
\[ \lim_{x \to \infty} \frac{\left( \sqrt{x^2 + 2x - 1} - x \right) \left( \sqrt{x^2 + 2x - 1} + x \right)}{\sqrt{x^2 + 2x - 1} + x} \]
This gives us:
\[ \lim_{x \to \infty} \frac{(\sqrt{x^2 + 2x - 1})^2 - x^2}{\sqrt{x^2 + 2x - 1} + x} \]

### Step 3: Simplify the Numerator
The numerator simplifies to:
\[ \lim_{x \to \infty} \frac{x^2 + 2x - 1 - x^2}{\sqrt{x^2 + 2x - 1} + x} = \lim_{x \to \infty} \frac{2x - 1}{\sqrt{x^2 + 2x - 1} + x} \]

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

### Step 5: Simplify the Limit
Now, we can simplify the limit:
\[ \lim_{x \to \infty} \frac{2 - \frac{1}{x}}{\sqrt{1 + \frac{2}{x} - \frac{1}{x^2}} + 1} \]

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

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