Let `f (x) =x + sqrt(x^(2) +2x) and g (x) = sqrt(x^(2) +2x)-x,` then:

Options

A. `lim _(x to oo) g (x) =1`
B. `lim _(x to oo) f (x)=1`
C. ` lim_(x to -oo) f (x)=-1`
D. `lim _(x to oo) g (x)=-1`

Answer

To solve the problem, we need to evaluate the limits of the functions \( f(x) \) and \( g(x) \) as \( x \) approaches infinity and negative infinity.

### Given Functions:
1. \( f(x) = x + \sqrt{x^2 + 2x} \)
2. \( g(x) = \sqrt{x^2 + 2x} - x \)

### Step 1: Evaluate \( \lim_{x \to \infty} g(x) \)

We start with the function \( g(x) \):
\[ g(x) = \sqrt{x^2 + 2x} - x \]

To simplify, we can rationalize the expression:
\[ g(x) = \frac{(\sqrt{x^2 + 2x} - x)(\sqrt{x^2 + 2x} + x)}{\sqrt{x^2 + 2x} + x} \]

This gives us:
\[ g(x) = \frac{x^2 + 2x - x^2}{\sqrt{x^2 + 2x} + x} = \frac{2x}{\sqrt{x^2 + 2x} + x} \]

Now, we can factor \( x \) out of the square root:
\[ g(x) = \frac{2x}{\sqrt{x^2(1 + \frac{2}{x})} + x} = \frac{2x}{x\sqrt{1 + \frac{2}{x}} + x} \]

This simplifies to:
\[ g(x) = \frac{2}{\sqrt{1 + \frac{2}{x}} + 1} \]

Now, taking the limit as \( x \to \infty \):
\[ \lim_{x \to \infty} g(x) = \frac{2}{\sqrt{1 + 0} + 1} = \frac{2}{1 + 1} = 1 \]

### Step 2: Evaluate \( \lim_{x \to -\infty} f(x) \)

Now we evaluate \( f(x) \):
\[ f(x) = x + \sqrt{x^2 + 2x} \]

We can again factor out \( x^2 \) from the square root:
\[ f(x) = x + \sqrt{x^2(1 + \frac{2}{x})} = x + |x|\sqrt{1 + \frac{2}{x}} \]

Since \( x \to -\infty \), \( |x| = -x \):
\[ f(x) = x - x\sqrt{1 + \frac{2}{x}} = x(1 - \sqrt{1 + \frac{2}{x}}) \]

Now, taking the limit as \( x \to -\infty \):
\[ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} x(1 - \sqrt{1 + 0}) = \lim_{x \to -\infty} x(1 - 1) = \lim_{x \to -\infty} x(0) = 0 \]

However, we need to analyze the expression more closely. As \( x \to -\infty \):
\[ f(x) = x(1 - \sqrt{1 + \frac{2}{x}}) = x(1 - (1 + \frac{1}{x} + O(\frac{1}{x^2}))) = x(-\frac{1}{x} + O(\frac{1}{x^2})) = -1 + O(\frac{1}{x}) \]
Thus, we find:
\[ \lim_{x \to -\infty} f(x) = -1 \]

### Conclusion:
- \( \lim_{x \to \infty} g(x) = 1 \)
- \( \lim_{x \to -\infty} f(x) = -1 \)

### Final Answers:
- \( \lim_{x \to \infty} g(x) = 1 \)
- \( \lim_{x \to -\infty} f(x) = -1 \)