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It looks like you're working on a limit problem from calculus. The expression seems to be:

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

You mentioned not using L'Hopital's Rule, which means we need to find another method to evaluate this limit.

### Solution without L'Hopital's Rule:
We can analyze the behavior of the numerator and denominator as \(x \to \infty\):

1. **Numerator**: The term \(2x\) grows linearly as \(x\) increases.
2. **Denominator**: The term \(e^x\) grows exponentially, which is much faster than \(x\), so \(e^x - x \approx e^x\) as \(x \to \infty\).

Thus, the fraction behaves like:
\[
\frac{2x}{e^x - x} \approx \frac{2x}{e^x}
\]
As \(x \to \infty\), the exponential growth of \(e^x\) dominates the linear term \(2x\), so:
\[
\lim_{x \to \infty} \frac{2x}{e^x} = 0
\]

Therefore, the limit is:
\[
\boxed{0}
\]

Let me know if you need more details or have further questions!

### Here are five related questions for practice:
1. What is the limit of \(\lim_{x \to \infty} \frac{x^2}{e^x}\)?
2. Evaluate \(\lim_{x \to \infty} \frac{3x^3}{e^x}\) without L'Hopital's rule.
3. Can you find \(\lim_{x \to \infty} \frac{e^x}{x^2}\)?
4. Analyze \(\lim_{x \to \infty} \frac{x}{x^2 + 1}\).
5. What is \(\lim_{x \to \infty} \frac{\ln(x)}{x}\)?

### Tip:
When dealing with limits involving exponential growth, remember that exponential functions like \(e^x\) grow much faster than any polynomial or logarithmic function.

Explore how to evaluate the limit \( \lim_{x \to \infty} \frac{2x}{e^x - x} \) without using L'Hopital's Rule. This problem illustrates exponential growth dominating polynomial growth in limit calculations, ideal for undergraduate calculus students.

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