Limits, Continuity & Differentiability Question - None

Question

$\lim_{n \to \infty} \frac{e^n}{\left(1 + \dfrac{1}{n}\right)^{n^2}}$ equals

Accepted Answer

Key Idea: Take log and use Taylor expansion of $\ln(1 + 1/n)$ to second term.

Step 1: Let $$L$$ be the limit. Then:
$\ln L = \lim_{n\to\infty}\left[n - n^2\ln\left(1+\frac{1}{n}\right)\right]$

Step 2: Expand $\ln(1+1/n)$ using Taylor series:
$\ln\left(1+\frac{1}{n}\right) = \frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \cdots$

Step 3: Multiply by $$n^2$$ :
$n^2\ln\left(1+\frac{1}{n}\right) = n - \frac{1}{2} + \frac{1}{3n} - \cdots$

Step 4: Substitute back:
$\ln L = \lim_{n\to\infty}\left[n - \left(n - \frac{1}{2} + \frac{1}{3n} - \cdots\right)\right] = \frac{1}{2}$

Step 5: Therefore $L = e^{1/2} = \sqrt{e}$ .

Answer: (D) $\sqrt{e}$

Trap: Students use

$(1+1/n)^n = e$

directly, writing

(1+1/n)^{n^2} = e^n

and getting

$L = 1$

. This ignores the error in the approximation —

$(1+1/n)^n$

approaches

$e$

from below, and that gap matters at

$n^2$

scale.

Key Concept: Taylor expansion of

\ln(1+x)

to second term for precise limit evaluation

Question: