lim⁡x→∞(5+ 3x/x^3 − 3−x/x+7 + 4+e^−x+x^3/ 2x^3+17x2+1)

We are tasked with evaluating the limit:

\[
\lim_{x \to \infty} \left( \frac{5 + 3x}{x^3} - \frac{3 - x}{x + 7} + \frac{4 + e^{-x} + x^3}{2x^3 + 17x^2 + 1} \right)
\]

To solve this limit, let's evaluate each term individually as \( x \to \infty \).

### First Term: \( \frac{5 + 3x}{x^3} \)
As \( x \to \infty \), this term can be simplified by dividing both the numerator and denominator by \( x^3 \):

\[
\frac{5 + 3x}{x^3} = \frac{5}{x^3} + \frac{3x}{x^3} = \frac{5}{x^3} + \frac{3}{x^2}
\]

Since \( \frac{5}{x^3} \to 0 \) and \( \frac{3}{x^2} \to 0 \) as \( x \to \infty \), we get:

\[
\lim_{x \to \infty} \frac{5 + 3x}{x^3} = 0
\]

### Second Term: \( \frac{3 - x}{x + 7} \)
As \( x \to \infty \), the dominant term in both the numerator and denominator is \( x \). We can approximate this term as:

\[
\frac{3 - x}{x + 7} \approx \frac{-x}{x} = -1
\]

Thus, the limit of this term is:

\[
\lim_{x \to \infty} \frac{3 - x}{x + 7} = -1
\]

### Third Term: \( \frac{4 + e^{-x} + x^3}{2x^3 + 17x^2 + 1} \)
As \( x \to \infty \), the dominant term in the numerator is \( x^3 \), and the dominant term in the denominator is also \( x^3 \). Therefore, we can simplify this term as:

\[
\frac{4 + e^{-x} + x^3}{2x^3 + 17x^2 + 1} \approx \frac{x^3}{2x^3} = \frac{1}{2}
\]

Thus, the limit of this term is:

\[
\lim_{x \to \infty} \frac{4 + e^{-x} + x^3}{2x^3 + 17x^2 + 1} = \frac{1}{2}
\]

### Combining the Results
Now, we combine the limits of the three terms:

\[
0 - 1 + \frac{1}{2} = -\frac{1}{2}
\]

Thus, the final answer is:

\[
\lim_{x \to \infty} \left( \frac{5 + 3x}{x^3} - \frac{3 - x}{x + 7} + \frac{4 + e^{-x} + x^3}{2x^3 + 17x^2 + 1} \right) = -\frac{1}{2}
\]

Would you like more details on any step?

Here are 5 related questions to further your understanding:
1. How do we determine which terms are dominant as \( x \to \infty \)?
2. What is the importance of simplifying terms when calculating limits?
3. How does the behavior of exponential terms like \( e^{-x} \) affect limits at infinity?
4. Can L'Hôpital's Rule be applied in cases like this? Why or why not?
5. How do rational functions behave as \( x \to \infty \)?

**Tip:** When working with limits at infinity, always compare the growth rates of the highest-degree terms in both the numerator and denominator.

Learn how to evaluate the limit of a complex rational function as x approaches infinity, covering key concepts such as asymptotic behavior, dominant term analysis, and simplification of rational expressions. Suitable for students in Grades 11-12.

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