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VITEEE 2019 Paper

Section: Mathematics

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Question : 93 of 125

Marks: +1, -0

Evaluate

\lim\limits_{x\to 2}\frac{\sqrt{x+7}-3\sqrt{2x-3}}{\sqrt[3]{x+6}-2\sqrt[3]{3x-5}}

$\frac{17}{9}$
$\frac{17}{18}$
$\frac{34}{23}$
$\frac{26}{7}$

Solution:

We have

L=\lim\limits_{x\to 2}\frac{\sqrt{x+7}-3\sqrt{2x-3}}{\sqrt[3]{x+6}-2\sqrt[3]{3x-5}}

(0/0\text{form})

Let

x-2=t

such that when x → 2, t → 0. Then

L=\lim\limits_{t\to 0}\frac{(t+9)^{1/2}-3(2t+1)^{1/2}}{(t+8)^{1/3}-2(3t+1)^{1/3}}

(0/0\text{form})

=\frac{3}{2}\lim\limits_{t\to 0}\frac{(1+t/9)^{1/2}-(2t+1)^{1/2}}{(1+t/8)^{1/3}-(3t+1)^{1/3}}

(0/0\text{form})

=\frac{3}{2}\lim\limits_{t\to 0}\frac{\frac{1}{2}\frac{t}{9}-(2t)\frac{1}{2}}{\frac{t}{8}\frac{1}{3}-(3t)\frac{1}{3}}

=\frac{3}{2}\frac{\frac{1}{18}-1}{\frac{1}{24}-1}=\frac{34}{23}

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