LimitshardFree

Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

The possible value(s) of

k

for which

\displaystyle\lim_{x\to\infty}\dfrac{2x^{3}-(\tan^{-1}x)^{3}}{\dfrac{8}{\pi}x^{3}\cot^{-1}(|kx|)+k^{2}x^{6}\sin\left(\dfrac{1}{x^{3}}\right)-3kx^{3}}=\dfrac{1}{2}

0

correct

-1

correct

2

correct

4

correct

Solution

Step 1: As

x\to\infty

: -

\tan^{-1}x\to\dfrac{\pi}{2}

, so

(\tan^{-1}x)^{3}\to\dfrac{\pi^{3}}{8}

, a constant. So numerator

\sim 2x^{3}

. -

\cot^{-1}(|kx|)\to 0

k\neq 0

(and if

k=0

\cot^{-1}0=\dfrac{\pi}{2}

). -

x^{6}\sin\dfrac{1}{x^{3}}\sim x^{6}\cdot\dfrac{1}{x^{3}}=x^{3}

. Step 2: Divide numerator and denominator by

x^{3}

. The numerator becomes

2-\dfrac{(\tan^{-1}x)^{3}}{x^{3}}\to 2-0=2

. Step 3: For the denominator (after dividing by

x^{3}

\dfrac{8}{\pi}\cot^{-1}(|kx|)+k^{2}x^{3}\sin\dfrac{1}{x^{3}}-3k

Step 4: Use

\cot^{-1}(|kx|)=\tan^{-1}\dfrac{1}{|kx|}\sim\dfrac{1}{|kx|}

for

k\neq 0

, so

\dfrac{8}{\pi}\cot^{-1}(|kx|)\to 0

. Also

k^{2}x^{3}\sin\dfrac{1}{x^{3}}\to k^{2}\cdot 1=k^{2}

. Step 5: So denominator

\to 0+k^{2}-3k=k^{2}-3k

. Step 6: Set the limit equal to

\dfrac{1}{2}

\dfrac{2}{k^{2}-3k}=\dfrac{1}{2}\;\Rightarrow\;k^{2}-3k=4\;\Rightarrow\;k^{2}-3k-4=0

(k-4)(k+1)=0\;\Rightarrow\;k=4\text{ or }k=-1

Step 7: Special case

k=0

. Then

\cot^{-1}(0)=\dfrac{\pi}{2}

, so

\dfrac{8}{\pi}\cdot\dfrac{\pi}{2}=4

. Also

k^{2}x^{6}\sin(\cdot)=0

and

-3k\cdot x^{3}=0

. After dividing by

x^{3}

: denominator

\to\dfrac{4}{1}\cdot

... need to recheck. Actually for

k=0

: numerator after

/x^{3}

2

, denominator after

/x^{3}

\dfrac{8}{\pi}\cdot\dfrac{\pi}{2}=4

. So limit

=\dfrac{2}{4}=\dfrac{1}{2}

. Valid. So

k=0

. Correct answers: (1)

0

, (2)

-1

and (4)

4

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