If `lim_(xto-oo)((3x^(4)+2x^(2)).sin(1/x)+|x|^(3)+5)/(|x|^(3)+|x^(2)|+|x|+1)=k` then `|k/2|` is …….

Answer

To solve the limit problem, we will follow these steps:

### Step 1: Analyze the limit expression
We need to evaluate the limit:
\[ \lim_{x \to -\infty} \frac{(3x^4 + 2x^2) \sin(1/x) + |x|^3 + 5}{|x|^3 + |x^2| + |x| + 1} \]

### Step 2: Simplify the absolute values
As \(x\) approaches \(-\infty\), we have \(|x| = -x\). Thus, we can rewrite the expression:
- In the numerator:
\[ |x|^3 = -x^3, \quad |x|^2 = -x^2 \]
- In the denominator:
\[ |x|^3 = -x^3, \quad |x|^2 = -x^2 \]

### Step 3: Substitute and simplify
Substituting these into the limit expression gives:
\[ \lim_{x \to -\infty} \frac{(3x^4 + 2x^2) \sin(1/x) - x^3 + 5}{-x^3 - x^2 - x + 1} \]

### Step 4: Evaluate \(\sin(1/x)\)
As \(x \to -\infty\), \(1/x \to 0\), and hence \(\sin(1/x) \to 0\). Therefore, the term \((3x^4 + 2x^2) \sin(1/x)\) approaches \(0\).

### Step 5: Focus on dominant terms
Now we focus on the dominant terms in both the numerator and denominator:
- Numerator: \(-x^3 + 5\)
- Denominator: \(-x^3 - x^2 - x + 1\)

### Step 6: Factor out \(-x^3\)
Factoring out \(-x^3\) from both the numerator and denominator:
\[ \lim_{x \to -\infty} \frac{-x^3(1 - \frac{5}{x^3})}{-x^3(1 + \frac{1}{x} + \frac{1}{x^2} - \frac{1}{x^3})} \]

### Step 7: Cancel out \(-x^3\)
Canceling \(-x^3\) from the numerator and denominator:
\[ \lim_{x \to -\infty} \frac{1 - \frac{5}{x^3}}{1 + \frac{1}{x} + \frac{1}{x^2} - \frac{1}{x^3}} \]

### Step 8: Evaluate the limit
As \(x \to -\infty\), the terms \(\frac{5}{x^3}\), \(\frac{1}{x}\), \(\frac{1}{x^2}\), and \(\frac{1}{x^3}\) all approach \(0\):
\[ \lim_{x \to -\infty} \frac{1 - 0}{1 + 0 + 0 - 0} = \frac{1}{1} = 1 \]

### Step 9: Find \(k\)
Thus, we find that \(k = 1\).

### Step 10: Calculate \(|k/2|\)
Now, we need to find \(|k/2|\):
\[ |k/2| = |1/2| = \frac{1}{2} \]

### Final Answer
The final answer is:
\[ \frac{1}{2} \]

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