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Limits Continuity and Differentiability

Section: Mathematics

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Question : 61 of 61

Marks: +1, -0

\lim\limits_{x\to\infty}\left(\;\frac{3x^2-2x+3}{3x^2+x-2}\right)^{3x-2}=

[AP EAPCET 18 May 2024 Shift 1]

$-3$
$e^{-1}$
$e^{-3}$
$-1$

Solution:

We want

\lim\limits_{x\to\infty}\left(\;\frac{3x^2-2x+3}{3x^2+x-2}\right)^{3x-2} .

Set

L = \left(\;\frac{3x^2-2x+3}{3x^2+x-2}\right)^{3x-2}, \qquad \text{ and take logs: }\quad \ln L = (3x-2)\ln\left(\;\frac{3x^2-2x+3}{3x^2+x-2}\right) .

First simplify inside:

\;\frac{3x^2-2x+3}{3x^2+x-2} = 1 + \;\frac{(3x^2-2x+3)-(3x^2+x-2)}{3x^2+x-2} = 1 + \;\frac{-3x+5}{3x^2+x-2} .

Call

u(x)=\;\frac{-3x+5}{3x^2+x-2}

so the base is

1+u(x)

and

u(x) \to 0

as

x \to \infty

.

For small

u, \ln(1+u) = u+O(u^2)

. Thus

\ln L=(3x-2)\ln(1+u(x)) \approx (3x-2)u(x) .

Compute the limit of

(3x-2)u(x)

:

(3x-2)u(x) = (3x-2)\cdot \;\frac{-3x+5}{3x^2+x-2} = \;\frac{(3x-2)(-3x+5)}{3x^2+x-2} .

Expand numerator:

(3x-2)(-3x+5) = -9x^2+21x-10

So

(3x-2)u(x) = \;\frac{-9x^2+21x-10}{3x^2+x-2} \sim \;\frac{-9x^2}{3x^2} = -3 \qquad \text{ as } x \to \infty .

More precisely, the limit is

\lim\limits_{x\to\infty}(3x-2)u(x) = -3

The extra

O(u^2)

term is of order

(3x-2)u(x)^2 \sim

const

\cdot x \cdot \;\frac{1}{x^2} \to 0

, so it doesn't affect the limit.

Hence

\lim\limits_{x\to\infty}\ln L = -3 \Rightarrow \qquad L = e^{-3}

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