The problem is to evaluate the following limit:

$\lim_{x \to 0} \frac{e^{-\frac{1}{x^2}}}{x^{100}}$

Solution:

1. Understanding the components of the expression:
   • $e^{-\frac{1}{x^2}}$: As $x \to 0^+$ or $x \to 0^-$, the term $-\frac{1}{x^2}$ tends to $-\infty$. Therefore, $e^{-\frac{1}{x^2}} \to 0$.
   • $x^{100}$: This term grows to $+\infty$ as $x \to 0^+$ and is undefined for $x \to 0^-$ due to the even power.

Thus, the numerator approaches $0$, while the denominator approaches $\infty$. This suggests that the overall fraction might approach $0$, but we need to formalize this further.

2. Substitution to simplify: Let $y = \frac{1}{x}$. Then, as $x \to 0^+$, $y \to +\infty$. Rewriting the limit in terms of $y$:

   $\lim_{x \to 0^+} \frac{e^{-\frac{1}{x^2}}}{x^{100}} = \lim_{y \to +\infty} \frac{e^{-y^2}}{\left(\frac{1}{y}\right)^{100}} = \lim_{y \to +\infty} y^{100} e^{-y^2}$

3. Analyze the new limit: We now need to evaluate $\lim_{y \to +\infty} y^{100} e^{-y^2}$.

   • $y^{100}$: This term grows polynomially (at a finite rate).
   • $e^{-y^2}$: This term decays exponentially fast, which dominates any polynomial growth.

   Therefore, the product $y^{100} e^{-y^2} \to 0$ as $y \to +\infty$.

4. Conclusion: $\lim_{x \to 0} \frac{e^{-\frac{1}{x^2}}}{x^{100}} = 0$

Final Answer:

$\boxed{0}$

Would you like a detailed explanation of any step, or do you have further questions? Here are some related problems you might want to explore:

1. What happens if the exponent in $x^{100}$ is changed to $x^n$, where $n$ is a smaller number?
2. How does the behavior of $e^{-1/x^2}$ differ for $x \to 0^+$ and $x \to 0^-$?
3. What if the numerator is $e^{1/x^2}$ instead of $e^{-1/x^2}$?
4. How does the comparison of exponential decay vs polynomial growth work in general limits?
5. Can this limit be solved using L'Hôpital's Rule, and how would that compare?

Tip: When dealing with limits involving exponentials and polynomials, exponential terms (growth or decay) usually dominate over polynomial terms!