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

Step 1: Set up the limit expression. Let the given limit be L. We have: L = _n (e^n)/((1 + 1)n)^n^2 We are given that L = sqrt(). Step 2: Simplify the expression by taking the natural logarithm. Let Y_n = (e^n)/((1 + 1)n)^n^2. Taking the natural logarithm of Y_n: (Y_n) = ((e^n)/((1 + 1)n)^n^2) Using logarithm properties ((A)/(B)) = A - B and (A^B) = B A: (Y_n) = (e^n) - ((1 + (1)/(n))^n^2) (Y_n) = n - n^2 (1 + (1)/(n)) Step 3: Evaluate the limit of (Y_n). We need to find _n [n - n^2 (1 + (1)/(n))]. Let x = (1)/(n). As n , x 0^+. Substitute x into the expression: _x 0^+ [(1)/(x) - (1)/(x^2) (1+x)] Combine the terms into a single fraction: _x 0^+ (x - (1+x))/(x^2) This is an indeterminate form of type (0)/(0) as x 0. We can use L'Hôpital's Rule or Taylor series expansion. Using Taylor series expansion for (1+x) around x=0: (1+x) = x - (x^2)/(2) + (x^3)/(3) - Substitute this into the limit expression: _x 0^+ (x - (x - x^2)/(2) + (x^3)/(3) - )x^2 _x 0^+ (x^2)/(2) - (x^3)/(3) + x^2 Divide each term by x^2: _x 0^+ ((1)/(2) - (x)/(3) + ) As x 0^+, all terms with x go to zero: _x 0^+ (Y_n) = (1)/(2) Step 4: Find the value of . Since _n (Y_n) = (1)/(2), we have: L = _n Y_n = e^1/2 = sqrt(e) We are given that L = sqrt(). Therefore, sqrt(e) = sqrt(). Squaring both sides gives: = e Step 5: Calculate []. The problem asks for [], which is the greatest integer function of . [] = [e] The value of e is approximately 2.71828. The greatest integer less than or equal to 2.71828 is 2. [e] = 2 The final answer is 2. Last free one today — make it count tomorrow, or type /upgrade for unlimited.