Question

Find ltx → ∞\((1+\frac{1}{x^2+2x+1})^{x^2+3x+1}\)

(a) e

(b) 1

(c) e^2

(d) ^1⁄e

I have been asked this question in a national level competition.

I'd like to ask this question from Indeterminate Forms topic in division Differential Calculus of Engineering Mathematics

Answer 1

Correct answer is (a) e

Easiest explanation: Use the form

lt x→ ∞(1 + f(x))^g(x) = e^lt x→ ∞ f(x) * g(x)

Provided as  x → ∞ we must have

f(x) → 0

g(x) → ∞

These conditions are met in our question

L = ltx → ∞\((1+\frac{1}{x^2+2x+1})^{x^2+3x+1} = e^{\frac{x^2+3x+1}{x^2+2x+1}}\)

ltx → 0\(\frac{x^2+3x+1}{x^2+2x+1}=1\)

L = e^1 = e.