`"lt"_(n to infty) (1-2+3-4 + 5-6+….-2n)/(sqrt(n^(2) + 1) + sqrt(4n^(2)+1))`=

Options

A. `-1//3`
B. `-1//5`
C. `1//3`
D. none

Answer

To solve the limit

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

we will break it down step by step.

### Step 1: Simplify the Numerator
The numerator \(1 - 2 + 3 - 4 + 5 - 6 + \ldots - 2n\) can be grouped into pairs:

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

Each pair sums to \(-1\). Since there are \(n\) pairs (from 1 to \(2n\)), the total sum of the numerator is:

\[ -n. \]

### Step 2: Simplify the Denominator
The denominator is

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

We can factor out \(n\) from each square root:

\[ \sqrt{n^2(1 + \frac{1}{n^2})} + \sqrt{4n^2(1 + \frac{1}{4n^2})} = n\sqrt{1 + \frac{1}{n^2}} + 2n\sqrt{1 + \frac{1}{4n^2}}. \]

Thus, the denominator simplifies to:

\[ n\left(\sqrt{1 + \frac{1}{n^2}} + 2\sqrt{1 + \frac{1}{4n^2}}\right). \]

### Step 3: Combine the Results
Now we can rewrite the limit:

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

The \(n\) in the numerator and denominator cancels out:

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

### Step 4: Evaluate the Limit
As \(n \to \infty\), both \(\frac{1}{n^2}\) and \(\frac{1}{4n^2}\) approach 0. Therefore, we have:

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

Thus, the limit becomes:

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

### Final Answer
The final answer is:

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