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

Answer

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

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

### Step 2: Rationalize the expression
To simplify \( \sqrt{x^2 + 1} - x \), we can multiply and divide by the conjugate \( \sqrt{x^2 + 1} + x \):
\[ \lim_{x \to \infty} 2x \cdot \frac{(\sqrt{x^2 + 1} - x)(\sqrt{x^2 + 1} + x)}{\sqrt{x^2 + 1} + x} \]
This gives us:
\[ \lim_{x \to \infty} 2x \cdot \frac{x^2 + 1 - x^2}{\sqrt{x^2 + 1} + x} = \lim_{x \to \infty} 2x \cdot \frac{1}{\sqrt{x^2 + 1} + x} \]

### Step 3: Simplify the expression
Now we simplify the limit:
\[ \lim_{x \to \infty} \frac{2x}{\sqrt{x^2 + 1} + x} \]

### Step 4: Factor out \( x \) from the square root
We can factor \( x \) out of the square root:
\[ \sqrt{x^2 + 1} = x\sqrt{1 + \frac{1}{x^2}} \]
Thus, the expression becomes:
\[ \lim_{x \to \infty} \frac{2x}{x(\sqrt{1 + \frac{1}{x^2}} + 1)} = \lim_{x \to \infty} \frac{2}{\sqrt{1 + \frac{1}{x^2}} + 1} \]

### Step 5: Evaluate the limit
As \( x \to \infty \), \( \frac{1}{x^2} \to 0 \):
\[ \lim_{x \to \infty} \frac{2}{\sqrt{1 + 0} + 1} = \frac{2}{1 + 1} = \frac{2}{2} = 1 \]

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