The problem in the image involves evaluating the limit:

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

Solution:

1. Rewrite the exponential term: Using the Taylor expansion for $e^{1/x}$ around $1/x \to 0$: $e^{1/x} = 1 + \frac{1}{x} + \frac{1}{2x^2} + \dots$

   So, the term becomes: $x e^{1/x} = x \left(1 + \frac{1}{x} + \frac{1}{2x^2} + \dots\right)$

2. Simplify the product: Expand the expression: $x e^{1/x} = x + 1 + \frac{1}{2x} + \dots$

3. Take the limit as $x \to \infty$:

   • The term $x$ dominates, as all other terms become negligible.
   • Therefore: $\lim_{x \to \infty} x e^{1/x} = \infty$

Conclusion:

The value of the limit is infinity.

Let me know if you'd like further clarification or a deeper breakdown!

Here are some follow-up questions for better understanding:

1. Why does the $x$ term dominate over others in the expansion?
2. What happens if we take $\lim_{x \to 0^+} x e^{1/x}$ instead?
3. How does the series expansion for $e^{1/x}$ help in solving this limit?
4. Can we solve this using L’Hôpital’s Rule? If so, how?
5. What happens if $e^{1/x}$ is replaced with $e^{-1/x}$?

Tip: When evaluating limits with exponential terms, expansions or logarithmic transformations can simplify