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 $x$.