Question

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

Answer 1

$\sqrt{e}$

Answer 2

To evaluate the limit $\lim_{n\to\infty} \frac{e^n}{(1+\frac{1}{n})^{n^2}}$, we can first take the natural logarithm of the expression.

Let $L = \lim_{n\to\infty} \frac{e^n}{(1+\frac{1}{n})^{n^2}}$.
Consider $\ln L$:

$$\ln L = \lim_{n\to\infty} \ln \left( \frac{e^n}{(1+\frac{1}{n})^{n^2}} \right)$$

Using logarithm properties, $\ln(a/b) = \ln a - \ln b$ and $\ln(x^y) = y \ln x$:

$$\ln L = \lim_{n\to\infty} \left[ \ln(e^n) - \ln\left((1+\frac{1}{n})^{n^2}\right) \right]$$ $$\ln L = \lim_{n\to\infty} \left[ n - n^2 \ln\left(1+\frac{1}{n}\right) \right]$$

This limit is of the indeterminate form $\infty - \infty$. To evaluate it, let $x = \frac{1}{n}$. As $n \to \infty$, $x \to 0$. Substituting $x$ into the expression:

$$\ln L = \lim_{x\to 0} \left[ \frac{1}{x} - \frac{1}{x^2} \ln(1+x) \right]$$

Combine the terms into a single fraction:

$$\ln L = \lim_{x\to 0} \frac{x - \ln(1+x)}{x^2}$$

This limit is now of the indeterminate form $\frac{0}{0}$. We can evaluate it using either L'Hopital's Rule or Taylor series expansion for $\ln(1+x)$.

Method 1: Using Taylor Series Expansion

The Taylor series expansion for $\ln(1+x)$ around $x=0$ is:

$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$$

Substitute this into the expression for $\ln L$:

$$\ln L = \lim_{x\to 0} \frac{x - \left(x - \frac{x^2}{2} + \frac{x^3}{3} - O(x^4)\right)}{x^2}$$ $$\ln L = \lim_{x\to 0} \frac{\frac{x^2}{2} - \frac{x^3}{3} + O(x^4)}{x^2}$$

Divide each term in the numerator by $x^2$:

$$\ln L = \lim_{x\to 0} \left( \frac{1}{2} - \frac{x}{3} + O(x^2) \right)$$

As $x \to 0$, all terms containing $x$ go to zero:

$$\ln L = \frac{1}{2}$$

Method 2: Using L'Hopital's Rule

Since the limit is of the form $\frac{0}{0}$, we can apply L'Hopital's Rule by differentiating the numerator and the denominator with respect to $x$:

$$\ln L = \lim_{x\to 0} \frac{\frac{d}{dx}(x - \ln(1+x))}{\frac{d}{dx}(x^2)}$$ $$\ln L = \lim_{x\to 0} \frac{1 - \frac{1}{1+x}}{2x}$$

Simplify the numerator:

$$\ln L = \lim_{x\to 0} \frac{\frac{(1+x)-1}{1+x}}{2x}$$ $$\ln L = \lim_{x\to 0} \frac{\frac{x}{1+x}}{2x}$$ $$\ln L = \lim_{x\to 0} \frac{x}{2x(1+x)}$$

Cancel out $x$ (since $x \ne 0$ in the limit process):

$$\ln L = \lim_{x\to 0} \frac{1}{2(1+x)}$$

Substitute $x=0$:

$$\ln L = \frac{1}{2(1+0)} = \frac{1}{2}$$

Both methods yield the same result for $\ln L$. Since $\ln L = \frac{1}{2}$, we can find $L$:

$$L = e^{1/2} = \sqrt{e}$$