`lim_(x rarr -oo)(1-2+3-4+5-6...-2x)/(sqrt(x^2+1)+sqrt(4x^2-1))`

Options

A. `1/3`
B. `-1/3`
C. `-1/5`
D. None of these

Answer

To solve the limit

\[ \lim_{x \to -\infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + \ldots - 2x}{\sqrt{x^2 + 1} + \sqrt{4x^2 - 1}}, \]

we will break it down step by step.

### Step 1: Simplifying the Numerator

The numerator can be expressed as:

\[ 1 - 2 + 3 - 4 + 5 - 6 + \ldots - 2x = (1 + 3 + 5 + \ldots + (2x - 1)) - (2 + 4 + 6 + \ldots + 2x). \]

The first part, \(1 + 3 + 5 + \ldots + (2x - 1)\), is the sum of the first \(x\) odd numbers, which is given by \(x^2\).

The second part, \(2 + 4 + 6 + \ldots + 2x\), is the sum of the first \(x\) even numbers, which is given by \(x(x + 1)\).

Thus, the numerator simplifies to:

\[ x^2 - x(x + 1) = x^2 - (x^2 + x) = -x. \]

### Step 2: Simplifying the Denominator

Now, we simplify the denominator:

\[ \sqrt{x^2 + 1} + \sqrt{4x^2 - 1}. \]

For large values of \(x\), we can factor out \(x^2\) from both square roots:

\[ \sqrt{x^2 + 1} = \sqrt{x^2(1 + \frac{1}{x^2})} = |x|\sqrt{1 + \frac{1}{x^2}}. \]

Since \(x \to -\infty\), \(|x| = -x\), so:

\[ \sqrt{x^2 + 1} = -x\sqrt{1 + \frac{1}{x^2}}. \]

Similarly, for the second term:

\[ \sqrt{4x^2 - 1} = \sqrt{4x^2(1 - \frac{1}{4x^2})} = 2|x|\sqrt{1 - \frac{1}{4x^2}} = -2x\sqrt{1 - \frac{1}{4x^2}}. \]

Thus, the denominator becomes:

\[ -x\sqrt{1 + \frac{1}{x^2}} - 2x\sqrt{1 - \frac{1}{4x^2}}. \]

### Step 3: Combining the Results

Now we can rewrite the limit:

\[ \lim_{x \to -\infty} \frac{-x}{-x\left(\sqrt{1 + \frac{1}{x^2}} + 2\sqrt{1 - \frac{1}{4x^2}}\right)}. \]

The \( -x \) terms cancel out:

\[ \lim_{x \to -\infty} \frac{1}{\sqrt{1 + \frac{1}{x^2}} + 2\sqrt{1 - \frac{1}{4x^2}}}. \]

### Step 4: Evaluating the Limit

As \(x\) approaches \(-\infty\), the terms \(\frac{1}{x^2}\) and \(\frac{1}{4x^2}\) approach \(0\):

\[ \sqrt{1 + \frac{1}{x^2}} \to 1 \quad \text{and} \quad 2\sqrt{1 - \frac{1}{4x^2}} \to 2. \]

Thus, we have:

\[ \lim_{x \to -\infty} \frac{1}{1 + 2} = \frac{1}{3}. \]

### Final Answer

Therefore, the limit is:

\[ \boxed{\frac{1}{3}}. \]