Let's solve the given limit:

$\lim_{x \to 0^+} \left( \frac{2}{e^x - 1} - \frac{1}{x^2} \right)$

Step-by-Step Solution:

Step 1: Analyze the individual terms as $x \to 0^+$

1. $\frac{2}{e^x - 1}$:

   • When $x \to 0^+$, $e^x - 1 \approx x$ (using the first-order approximation of $e^x$).
   • So, $\frac{2}{e^x - 1} \approx \frac{2}{x}$.

2. $\frac{1}{x^2}$:

   • As $x \to 0^+$, this term diverges to $+\infty$.

Step 2: Combine the terms

The limit becomes: $\lim_{x \to 0^+} \left( \frac{2}{x} - \frac{1}{x^2} \right)$

Step 3: Rewrite with a common denominator

$\frac{2}{x} - \frac{1}{x^2} = \frac{2x - 1}{x^2}.$

Step 4: Evaluate the numerator and denominator as $x \to 0^+$

• The numerator $2x - 1 \to -1$ as $x \to 0^+$.
• The denominator $x^2 \to 0^+$, making the fraction diverge to $-\infty$.

Conclusion:

$\lim_{x \to 0^+} \left( \frac{2}{e^x - 1} - \frac{1}{x^2} \right) = -\infty.$

Would you like a detailed breakdown of any specific step? Here are 5 questions to deepen your understanding:

1. How do Taylor expansions help in approximating $e^x - 1$?
2. Why does $x^2 \to 0$ dominate in the denominator as $x \to 0$?
3. Could you explain the behavior of each term as $x \to 0^+$ separately?
4. How does the negative sign in $2x - 1$ influence the limit's outcome?
5. What would change in the calculation if $x \to 0^-$?

Tip: When solving limits, substituting approximations like Taylor series for small values of $x$ can simplify the analysis dramatically.