If displaystyle limx →∞ (1+ (a/x) + (b/x2))2x = e2 , then the values of a and b, are

Question

If displaystyle limx →∞ (1+ (a/x) + (b/x2))2x = e2 , then the values of a and b, are (A) a = 1 and b = 2 (B) a = 1, b ∈ R (C) a ∈ R, b = 2 (D) a ∈ R, b ∈ R

Tardigrade Admin · Accepted Answer

We know that displaystyle limx→∞ (1+x)(1/x) =e ∴ displaystyle limx→∞ (1+ (a/x) + (b/x2))2x = e2 ⇒ displaystyle limx→∞ [(1+ (a/x) + (b/x2)) ((1/(a)x+ (b)x2)]2x((a/x) + (b/x2)) =e2 ⇒ e displaystyle lim/x→∞)2[a+ (b/x)] = e2 ⇒ e2a = e2 ⇒ a=1 , b ∈ R

Q. If $x \to \infty lim (1 + \frac{a}{x} + \frac{b}{x ^{2}})^{2 x} = e^{2}$ , then the values of a and b, are

$a = 1$ and $b = 2$

$a = 1, b \in R$

$a \in R, b = 2$

$a \in R, b \in R$

Q. If x→∞lim​(1+xa​+x2b​)2x=e2 , then the values of a and b, are

a=1 and b=2

a=1,b∈R

a∈R,b=2

a∈R,b∈R

We know that x→∞lim​(1+x)x1​=e ∴x→∞lim​(1+xa​+x2b​)2x=e2 ⇒x→∞lim​⎣⎡​(1+xa​+x2b​)(xa​+x2b​1​)⎦⎤​2x(xa​+x2b​)=e2 ⇒ex→∞lim​2[a+xb​]=e2 ⇒e2a=e2 ⇒a=1 , b∈R

Q. If $x \to \infty lim (1 + \frac{a}{x} + \frac{b}{x ^{2}})^{2 x} = e^{2}$ , then the values of a and b, are

$a = 1$ and $b = 2$

$a = 1, b \in R$

$a \in R, b = 2$

$a \in R, b \in R$