The value of `lim_(xrarr-oo)(x^(2)tan((2)/(x)))/(sqrt(16x^(2)-x+1))` is equal to

Options

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

Answer

To solve the limit

\[ \lim_{x \to -\infty} \frac{x^2 \tan\left(\frac{2}{x}\right)}{\sqrt{16x^2 - x + 1}}, \]

we will follow these steps:

### Step 1: Analyze the numerator

As \( x \to -\infty \), the term \( \frac{2}{x} \) approaches \( 0 \). We can use the small angle approximation for the tangent function:

\[ \tan\left(\frac{2}{x}\right) \approx \frac{2}{x} \quad \text{(for small values of } \frac{2}{x}). \]

Thus, the numerator becomes:

\[ x^2 \tan\left(\frac{2}{x}\right) \approx x^2 \cdot \frac{2}{x} = 2x. \]

### Step 2: Analyze the denominator

Next, we simplify the denominator:

\[ \sqrt{16x^2 - x + 1}. \]

As \( x \to -\infty \), the dominant term in the square root is \( 16x^2 \). Therefore, we can factor out \( x^2 \):

\[ \sqrt{16x^2 - x + 1} = \sqrt{x^2(16 - \frac{1}{x} + \frac{1}{x^2})} = |x|\sqrt{16 - \frac{1}{x} + \frac{1}{x^2}}. \]

Since \( x \) is negative as \( x \to -\infty \), we have \( |x| = -x \). Thus, the denominator simplifies to:

\[ \sqrt{16x^2 - x + 1} \approx -x \sqrt{16} = -4x. \]

### Step 3: Substitute back into the limit

Now we can substitute our approximations back into the limit:

\[ \lim_{x \to -\infty} \frac{2x}{-4x} = \lim_{x \to -\infty} \frac{2}{-4} = -\frac{1}{2}. \]

### Step 4: Conclusion

Thus, the value of the limit is:

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