Lim n eⁿ / (1 + 1/n)ⁿ2 = sqrt(lambda), then [lambda] is, [.] -> greatest integer function

Let the given limit be $L$. We have:

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

We are given that $L = \sqrt{\lambda}$.

Let $Y_n = \frac{e^n}{\left(1 + \frac{1}{n}\right)^{n^2}}$.
Taking the natural logarithm of $Y_n$:

$$\ln(Y_n) = \ln\left(\frac{e^n}{\left(1 + \frac{1}{n}\right)^{n^2}}\right)$$

Using logarithm properties $\ln\left(\frac{A}{B}\right) = \ln A - \ln B$ and $\ln(A^B) = B \ln A$:

$$\ln(Y_n) = \ln(e^n) - \ln\left(\left(1 + \frac{1}{n}\right)^{n^2}\right)$$ $$\ln(Y_n) = n - n^2 \ln\left(1 + \frac{1}{n}\right)$$

We need to find $\lim_{n \to \infty} \left[n - n^2 \ln\left(1 + \frac{1}{n}\right)\right]$.
Let $x = \frac{1}{n}$. As $n \to \infty$, $x \to 0^+$.
Substitute $x$ into the expression:

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

Combine the terms into a single fraction:

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

This is an indeterminate form of type $\frac{0}{0}$ as $x \to 0$. We can use L'Hôpital's Rule or Taylor series expansion.

Using Taylor series expansion for $\ln(1+x)$ around $x=0$:
$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$
Substitute this into the limit expression:

$$\lim_{x \to 0^+} \frac{x - \left(x - \frac{x^2}{2} + \frac{x^3}{3} - \dots\right)}{x^2}$$ $$\lim_{x \to 0^+} \frac{\frac{x^2}{2} - \frac{x^3}{3} + \dots}{x^2}$$

Divide each term by $x^2$:

$$\lim_{x \to 0^+} \left(\frac{1}{2} - \frac{x}{3} + \dots\right)$$

As $x \to 0^+$, all terms with $x$ go to zero:

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

Since $\lim_{n \to \infty} \ln(Y_n) = \frac{1}{2}$, we have:

$$L = \lim_{n \to \infty} Y_n = e^{1/2} = \sqrt{e}$$

We are given that $L = \sqrt{\lambda}$.
Therefore, $\sqrt{e} = \sqrt{\lambda}$.
Squaring both sides gives:

$$\lambda = e$$

The problem asks for $[\lambda]$, which is the greatest integer function of $\lambda$.

$$[\lambda] = [e]$$

The value of $e$ is approximately 2.71828.
The greatest integer less than or equal to 2.71828 is 2.

$$[e] = 2$$