https://file.aiquickdraw.com/mathbot/file/1726759442459d32m08pg.jpg

The expression in the image is:

\[
\lim_{x \to -\infty} \frac{2x}{e^x - x}
\]

To evaluate this limit, let's analyze the behavior of both the numerator and the denominator as \( x \) approaches \(-\infty\).

1. **Numerator**: \( 2x \) as \( x \to -\infty \) will tend towards \(-\infty\).

2. **Denominator**: \( e^x - x \) as \( x \to -\infty \):
    - The exponential function \( e^x \) tends to \( 0 \) very rapidly because exponential decay is very fast as \( x \) becomes large and negative.
    - The linear term \(-x\) tends to \( +\infty \) (since \( x \) is negative, \(-x\) is positive and large).

So, as \( x \) approaches \(-\infty\), the denominator \( e^x - x \) behaves approximately like \(-x\), because \( -x \) dominates \( e^x \).

Thus, the limit simplifies to:

\[
\lim_{x \to -\infty} \frac{2x}{e^x - x} \approx \lim_{x \to -\infty} \frac{2x}{-x} = \lim_{x \to -\infty} -2 = -2
\]

So the limit is:

\[
\lim_{x \to -\infty} \frac{2x}{e^x - x} = -2
\]

Would you like more details on any part of this calculation?

Here are five related questions to expand on this concept:

1. How would the limit change if the expression were \( \lim_{x \to -\infty} \frac{x^2}{e^x - x} \)?
2. What happens to the limit if the numerator was \( e^x \) instead of \( 2x \)?
3. Can L'Hôpital's Rule be applied to solve this limit, and how would the process differ?
4. How does the behavior of exponential functions compare with polynomial functions as \( x \) tends to \(-\infty\)?
5. How would you evaluate \( \lim_{x \to \infty} \frac{2x}{e^x - x} \)?

**Tip:** When dealing with limits involving exponential functions, always consider the dominant terms in the numerator and denominator, especially for large positive or negative values of \( x \).

Evaluate the limit: \( \lim_{x \to -\infty} \frac{2x}{e^x - x} \).

Explore the evaluation of the limit \( \lim_{x \to -\infty} \frac{2x}{e^x - x} \) and understand how exponential and linear functions behave at infinity. This solution breaks down the asymptotic dominance between terms and provides step-by-step guidance, suitable for advanced high school math students.

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