`lim_(x->-oo)(x^5tan(1/(pix^2))+3|x^2|+7)/(|x^3|+7|x|+8)` is equal to: a. `-1/pi` b. 0 c. `oo` d. Does not exist

Options

A. `(-1)/(pi)`
B. `0`
C. `oo`
D. Does not exist

Answer

To solve the limit

\[ \lim_{x \to -\infty} \frac{x^5 \tan\left(\frac{1}{\pi x^2}\right) + 3|x^2| + 7}{|x^3| + 7|x| + 8}, \]

we will follow these steps:

### Step 1: Simplify the absolute values
Since \( x \to -\infty \), we have:
- \( |x| = -x \)
- \( |x^2| = x^2 \)
- \( |x^3| = -x^3 \)

Thus, we can rewrite the limit as:

\[ \lim_{x \to -\infty} \frac{x^5 \tan\left(\frac{1}{\pi x^2}\right) + 3x^2 + 7}{-x^3 - 7x + 8}. \]

### Step 2: Analyze \(\tan\left(\frac{1}{\pi x^2}\right)\)
As \( x \to -\infty \), \( \frac{1}{\pi x^2} \to 0 \). We know that \(\tan(z) \approx z\) when \( z \) is close to 0. Therefore,

\[ \tan\left(\frac{1}{\pi x^2}\right) \approx \frac{1}{\pi x^2}. \]

### Step 3: Substitute back into the limit
Substituting this approximation into our limit gives:

\[ \lim_{x \to -\infty} \frac{x^5 \cdot \frac{1}{\pi x^2} + 3x^2 + 7}{-x^3 - 7x + 8}. \]

This simplifies to:

\[ \lim_{x \to -\infty} \frac{\frac{x^5}{\pi x^2} + 3x^2 + 7}{-x^3 - 7x + 8} = \lim_{x \to -\infty} \frac{\frac{x^3}{\pi} + 3x^2 + 7}{-x^3 - 7x + 8}. \]

### Step 4: Divide numerator and denominator by \( x^3 \)
Now, we divide both the numerator and the denominator by \( x^3 \):

\[ \lim_{x \to -\infty} \frac{\frac{1}{\pi} + \frac{3}{x} + \frac{7}{x^3}}{-1 - \frac{7}{x^2} + \frac{8}{x^3}}. \]

### Step 5: Evaluate the limit
As \( x \to -\infty \), the terms \( \frac{3}{x} \), \( \frac{7}{x^3} \), \( \frac{7}{x^2} \), and \( \frac{8}{x^3} \) all approach 0. Thus, we have:

\[ \lim_{x \to -\infty} \frac{\frac{1}{\pi} + 0 + 0}{-1 + 0 + 0} = \frac{\frac{1}{\pi}}{-1} = -\frac{1}{\pi}. \]

### Final Answer
Thus, the limit is:

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