We are asked to compute the limit:

$\lim_{x \to +\infty} \frac{e^{1-x}}{1/x}$

Step-by-step Solution:

1. Rewrite the expression:

   We can rewrite the limit as:

   $\lim_{x \to +\infty} \frac{e^{1-x}}{1/x} = \lim_{x \to +\infty} e^{1-x} \cdot x$

2. Simplify the exponent:

   The expression $e^{1-x}$ can be rewritten as:

   $e^{1-x} = e \cdot e^{-x}$

   Thus, the limit becomes:

   $\lim_{x \to +\infty} x \cdot e \cdot e^{-x}$

   or simply:

   $e \lim_{x \to +\infty} x e^{-x}$

3. Evaluate the limit:

   To compute $\lim_{x \to +\infty} x e^{-x}$, notice that $e^{-x}$ decays exponentially, while $x$ grows linearly. Exponentially decaying functions approach zero much faster than linear growth, so this product tends to 0:

   $\lim_{x \to +\infty} x e^{-x} = 0$

4. Final Answer:

   Therefore, multiplying by $e$, we get:

   $\lim_{x \to +\infty} e^{1-x} \cdot x = e \cdot 0 = 0$

Conclusion:

$\lim_{x \to +\infty} \frac{e^{1-x}}{1/x} = 0$

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. What is the limit of $\lim_{x \to +\infty} \frac{x^2}{e^x}$?
2. How do we apply L'Hopital's Rule to indeterminate forms in limits?
3. How does the behavior of $e^x$ compare to polynomial functions as $x \to +\infty$?
4. Can you explain the comparison between exponential growth and linear growth in limits?
5. How can we approximate limits involving exponential and logarithmic functions?

Tip: Exponentially decaying functions like $e^{-x}$ tend to dominate polynomial terms as $x \to +\infty$, leading the whole expression to approach zero.