The value of `lim_(xrarr-oo)(x^(2)tan((1)/(x)))/(sqrt(4x^(2)-x+1))` is equal to

Options

A. `1`
B. `(1)/(2)`
C. `-1`
D. `-(1)/(2)`

Answer

To find the limit

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

we can follow these steps:

### Step 1: Substitute \( x = \frac{1}{n} \)

As \( x \to -\infty \), we can substitute \( x = -n \) where \( n \to \infty \). Therefore, we rewrite the limit as:

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

### Step 2: Simplify the expression

Now, we simplify the expression:

1. The numerator becomes:
\[ n^2 \tan\left(-\frac{1}{n}\right) = -n^2 \tan\left(\frac{1}{n}\right). \]
Using the small angle approximation \( \tan(x) \approx x \) for small \( x \), we have:
\[ \tan\left(\frac{1}{n}\right) \approx \frac{1}{n}. \]
Thus, the numerator simplifies to:
\[ -n^2 \cdot \frac{1}{n} = -n. \]

2. The denominator becomes:
\[ \sqrt{4n^2 + n + 1}. \]
For large \( n \), this can be approximated as:
\[ \sqrt{4n^2} = 2n. \]

### Step 3: Substitute back into the limit

Now substituting back into the limit, we have:

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

### Conclusion

Thus, the value of the limit is:

\[ \boxed{-\frac{1}{2}}. \]