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Limits Continuity and Differentiability question

2004 · Shift 0 · Q72

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JEE Main
Mathematics
Limits Continuity and Differentiability
2004 · Shift 0 · Q72

Limits Continuity and Differentiability question

2004 · Shift 0 · Q72

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

If

\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)^{2x}} = {e^2}

, then the value of

a

and

b

, are

A
$a$ = 1 and $b$ = 2
B
$a$ = 1 and $b \in R$
C
$a \in R$ and $b$ = 2
D
$a \in R$ and $b \in R$

View written solutionFree

Correct answer: B

We need to evaluate

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

Use the standard exponential-limit idea: If

L=\lim_{x\to\infty}\left(1+u_x\right)^{v_x},

with $u_x\to 0$ , then

\ln L=\lim_{x\to\infty} v_x\ln(1+u_x).

Here,

u_x=\frac{a}{x}+\frac{b}{x^2}, \qquad v_x=2x.

So

\ln L=\lim_{x\to\infty}2x\ln\left(1+\frac{a}{x}+\frac{b}{x^2}\right).

Expand $\ln(1+t)$ for small $t$ :

\ln(1+t)=t-\frac{t^2}{2}+o(t^2).

Take

t=\frac{a}{x}+\frac{b}{x^2}.

Then

\ln\left(1+\frac{a}{x}+\frac{b}{x^2}\right) =\frac{a}{x}+\frac{b}{x^2}-\frac12\left(\frac{a}{x}+\frac{b}{x^2}\right)^2+o\left(\frac1{x^2}\right).

For the limit, only the leading term matters after multiplying by $2x$ :

2x\ln\left(1+\frac{a}{x}+\frac{b}{x^2}\right) =2x\left(\frac{a}{x}+O\left(\frac1{x^2}\right)\right) =2a+O\left(\frac1x\right).

Hence,

\ln L=2a.

Therefore,

L=e^{2a}.

Given that $L=e^2$ , we get

e^{2a}=e^2 \implies 2a=2 \implies a=1.

Observe that $b$ does not affect the limit, because its contribution is of order $\frac1x$ in the exponent and vanishes as $x\to\infty$ . So

b\in \mathbb R.

Check options:

A: $a=1, b=2$ — too restrictive.
B: $a=1, b\in\mathbb R$ — correct.
C: $a\in\mathbb R, b=2$ — wrong.
D: $a\in\mathbb R, b\in\mathbb R$ — wrong.

Therefore, the correct option is B.

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