Question

Let \( x_n = n^{1/n} \) and \( y_n = e^{1 - x_n}, \, n \in \mathbb{N}. \) Then the value of \( \lim_{n \to \infty} y_n \) is ..............

Hint:

When dealing with \( n^{1/n} \), remember that it tends to 1 as \( n \to \infty \). Use logarithmic expansion for precision.

Answer 1

Correct Answer: 1

Step 1: Find \( \lim_{n \to \infty} x_n. \)
\[ x_n = n^{1/n} = e^{\frac{\ln n}{n}}. \] As \( n \to \infty \), \( \frac{\ln n}{n} \to 0 \), so \( x_n \to e^0 = 1. \)

Step 2: Compute \( \lim_{n \to \infty} y_n. \)
\[ y_n = e^{1 - x_n} \to e^{1 - 1} = e^0 = 1. \] Wait—correction! We must find the limiting value more carefully. Since \( x_n \approx 1 + \frac{\ln n}{n} \) for large \( n \), \[ 1 - x_n \approx -\frac{\ln n}{n}. \] Then \[ y_n = e^{1 - x_n} = e^{-\frac{\ln n}{n}} = (e^{\ln n})^{-1/n} = n^{-1/n} \to 1. \] But we need to check scaling with the definition \( y_n = e^{1 - x_n} \). Substituting \( x_n \to 1 \), \[ \lim_{n \to \infty} y_n = e^{1 - 1} = 1. \] Hence, \( \boxed{1} \).

Final Answer: \[ \boxed{1} \]