_ x → a (a + 2x)(1)/(3) -(3x)(1)/(3) (3a + x\ right)(1)/(3) -(4x) dfrac 1 3 (a ≠ 0) & is equal to:

Correct answer: D. ((2)/(3)) ((2)/(9))(1)/(3) At x = a , limit is of (0)/(0) form. So, applying L'Hospital rule, we get _ x → a (a + 2x) ^ (1)/(3) - (3x)(1)/(3) (3a + x)(1)/(3) - (4x)(1)/(3) = _ x → a (1)/(3) (a + 2x) ^ - (2)/(3) . 2 - (1)/(3) . (3x) ^ - (2)/(3) . 3 (1)/(3) (3a + x) ^ - (2)/(3) . - (1)/(3) . (4x) ^ - (2)/(3) . 4 = (1)/(3) (3a) ^ - (2)/(3) . (2 - 3) (1)/(3) (4a) ^ - (2)/(3) . (1 - 4) = 3 ^ - (2)/(3) 4 ^ - (2)/(3) . (1)/(3) = (2(4)/(3))/(9(1)/(3)) . (1)/(3) = ((2)/(3))((2)/(9))(1)/(3)

JEE Main 20203 SeptemberEvening Shift

Exam: JEE Main
Year: 2020
Date: 3 September
Shift: Evening Shift
Subject: Mathematics
Chapter: Limits

JEE Main 2020 · 3 September · Evening Shift · Mathematics · Limits

_ x → a (a + 2x)(1)/(3) -(3x)(1)/(3) (3a + x\ right)(1)/(3) -(4x)...

\lim_{x \to a}\dfrac{

\left(a + 2x\right)\dfrac{1}{3} -\left(3x\right)\dfrac{1}{3}}{ \left(3a + x\$

right)\dfrac{1}{3} -\left(4x\right)\d

dfrac{1}{3}}\left(a \ne 0\right)$& is equal to:

Options

A\left(\dfrac{2}{9}\right) $\left(\dfrac{2}{3}\right)\dfrac{1}{3}$
B $\left(\dfrac{2}{3}\right)\dfrac{4}{3}$
C $\left(\dfrac{2}{9}\right)\dfrac{4}{3}$
D\left(\dfrac{2}{3}\right) $\left(\dfrac{2}{9}\right)\dfrac{1}{3}$

Correct answer

D. \left(\dfrac{2}{3}\right) $\left(\dfrac{2}{9}\right)\dfrac{1}{3}$

Step-by-step solution

At $x = a$ , limit is of $\dfrac{0}{0}$ form.

So, applying L'Hospital rule, we get

$\lim_{x \to a}\dfrac{{\left(a + 2x\right)}^{\dfrac{1}{3}} - \left(3x\right)\dfrac{1}{3}}{\left(3a + x\right)\dfrac{1}{3} - \left(4x\right)\dfrac{1}{3}} = \lim_{x \to a}\dfrac{\dfrac{1}{3}{\left(a + 2x\right)}^{- \dfrac{2}{3}} . 2 - \dfrac{1}{3} . {\left(3x\right)}^{- \dfrac{2}{3}} . 3}{\dfrac{1}{3}{\left(3a + x\right)}^{- \dfrac{2}{3}} . - \dfrac{1}{3} . {\left(4x\right)}^{- \dfrac{2}{3}} . 4}$

$= \dfrac{\dfrac{1}{3}{\left(3a\right)}^{- \dfrac{2}{3}} . \left(2 - 3\right)}{\dfrac{1}{3}{\left(4a\right)}^{- \dfrac{2}{3}} . \left(1 - 4\right)} = \dfrac{{3}^{- \dfrac{2}{3}}}{{4}^{- \dfrac{2}{3}}} . \dfrac{1}{3}$

$= \dfrac{2\dfrac{4}{3}}{9\dfrac{1}{3}} . \dfrac{1}{3} = \left(\dfrac{2}{3}\right)\left(\dfrac{2}{9}\right)\dfrac{1}{3}$

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