If $\lim _{x \rightarrow \infty}\left(1+\frac{a}{x}+\frac{b}{x^{2}}\right)^{2 x}=e^{2}$, then the values of $a$ and $b$, are

(B) $a=1, b \in \underline{\underline{R}}$

(A) $a \in \underline{\underline{R}}, b \in \underline{\underline{R}}$

(C) $a \in \underline{\underline{R}}, b=2$

JEE Main · 2004 · 24 Apr 2004 · Offline · Limits

lim_{x \to \infty} (1 + \frac{a}{x} + \frac{b}{x ^{2}})^{2 x} = e^{2}

, then the values of

a

and

b

, are

Answer: (B)

Factor the exponent as

(1 + \frac{a}{x} + \frac{b}{x ^{2}})^{2 x} = [(1 + \frac{a}{x} + \frac{b}{x ^{2}})^{x}]^{2}

. As

x \to \infty

the

\frac{b}{x ^{2}}

piece is negligible beside

\frac{a}{x}

, so the inner limit is

e^{a}

and the whole expression tends to

e^{2 a}

. Matching

e^{2 a} = e^{2}

forces

a = 1

, while

b

never enters the leading behaviour, so

b

may be any real number.

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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