Question

Which of the following assertions is correct? \[\text{(I) } \lim_{n\to\infty}\left(1+\frac1n\right)^{n^2}e^{-n} = 1 \qquad \text{(II) For } 0 < a < 1 < b,\ \lim_{n\to\infty}\left(a+\frac{b}{n}\right)^n = e^{b/a}\]

• A. Only (I)
• B. Only (II)
• C. Both (I) and (II)
• D. Neither (I) nor (II)

Hint:

Expand ln(1+1/n) for (I); for (II) note a^n to 0 since 0<a<1 dominates the finite factor.

Answer 1

The Correct Option is D

Step 1: Evaluate the limit in (I). Take logarithms: \[\ln\left[(1+1/n)^{n^2}\right] = n^2\ln\left(1+\frac1n\right).\] Using \(\ln(1+t) = t - \frac{t^2}{2}+\frac{t^3}{3}-\cdots\) with \(t=1/n\), \[n^2\ln\left(1+\frac1n\right) = n^2\left(\frac1n - \frac{1}{2n^2}+\frac{1}{3n^3}-\cdots\right) = n - \frac12 + \frac{1}{3n} - \cdots\]

Step 2: Combine with the \(e^{-n}\) factor. \[(1+1/n)^{n^2}e^{-n} = \exp\left(n - \frac12 + O\!\left(\frac1n\right) - n\right) = \exp\left(-\frac12 + O\!\left(\frac1n\right)\right) \longrightarrow e^{-1/2}.\] Since \(e^{-1/2} = 1/\sqrt e \ne 1\), statement (I) is FALSE.

Step 3: Evaluate the limit in (II). Write \[\left(a+\frac bn\right)^n = a^n\left(1+\frac{b}{an}\right)^n.\] As \(n\to\infty\), \(\left(1+\dfrac{b}{an}\right)^n \to e^{b/a}\), a finite number. But since \(0<a<1\), \(a^n \to 0\).

Step 4: Combine. \[\left(a+\frac bn\right)^n = a^n\left(1+\frac{b}{an}\right)^n \longrightarrow 0 \cdot e^{b/a} = 0.\] So the limit is \(0\), not \(e^{b/a}\); statement (II) is also FALSE.

Step 5: Conclusion. Neither (I) nor (II) is correct.

\[\boxed{\text{Neither (I) nor (II)}}\]