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QJEEM21Ma6550: JEE Main 2021 Mathematics - Class 12 JEEM21Ma6550 Solution: The value of lim x → 0 ⁡ ( x 1 − sin ⁡ x 8 − 1 + sin ⁡ x 8 ) is …

The value of

limx→0⁡(x1−sin⁡x8−1+sin⁡x8) is equal to :

Multiple Choice Options

  • A 0
  • B 4
  • C −- 4
  • D −- 1

Step-by-Step Verified Solution

limx→0⁡(x1−sin⁡x8−1+sin⁡x8)

= lim⁡x→0x[(1−sin⁡x)18−(1+sin⁡x)18]×[(1−sin⁡x)18+(1+sin⁡x)18(1−sin⁡x)18+(1+sin⁡x)18]\mathop {\lim }\limits_{x \to 0} {x \over {\left[ {{{\left( {1 - \sin x} \right)}^{{1 \over 8}}} - {{\left( {1 + \sin x} \right)}^{{1 \over 8}}}} \right]}} \times \left[ {{{{{\left( {1 - \sin x} \right)}^{{1 \over 8}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 8}}}} \over {{{\left( {1 - \sin x} \right)}^{{1 \over 8}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 8}}}}}} \right]

= lim⁡x→0x[(1−sin⁡x)18+(1+sin⁡x)18][(1−sin⁡x)14−(1+sin⁡x)14]×[(1−sin⁡x)14+(1+sin⁡x)14(1−sin⁡x)14+(1+sin⁡x)14]\mathop {\lim }\limits_{x \to 0} {{x\left[ {{{\left( {1 - \sin x} \right)}^{{1 \over 8}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 8}}}} \right]} \over {\left[ {{{\left( {1 - \sin x} \right)}^{{1 \over 4}}} - {{\left( {1 + \sin x} \right)}^{{1 \over 4}}}} \right]}} \times \left[ {{{{{\left( {1 - \sin x} \right)}^{{1 \over 4}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 4}}}} \over {{{\left( {1 - \sin x} \right)}^{{1 \over 4}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 4}}}}}} \right]

= lim⁡x→0x[2][2][(1−sin⁡x)12−(1+sin⁡x)12]×[(1−sin⁡x)12+(1+sin⁡x)12(1−sin⁡x)12+(1+sin⁡x)12]\mathop {\lim }\limits_{x \to 0} {{x\left[ 2 \right]\left[ 2 \right]} \over {\left[ {{{\left( {1 - \sin x} \right)}^{{1 \over 2}}} - {{\left( {1 + \sin x} \right)}^{{1 \over 2}}}} \right]}} \times \left[ {{{{{\left( {1 - \sin x} \right)}^{{1 \over 2}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 2}}}} \over {{{\left( {1 - \sin x} \right)}^{{1 \over 2}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 2}}}}}} \right]

= lim⁡x→0x[2][2][2][(1−sin⁡x)−(1+sin⁡x)]\mathop {\lim }\limits_{x \to 0} {{x\left[ 2 \right]\left[ 2 \right]\left[ 2 \right]} \over {\left[ {\left( {1 - \sin x} \right) - \left( {1 + \sin x} \right)} \right]}}

=lim⁡x→0(−12)(2)(2)(2)= \mathop {\lim }\limits_{x \to 0} \left( { - {1 \over 2}} \right)(2)(2)(2)

= -4

∵\because {lim⁡x→0sin⁡xx=1}\left\{ {\mathop {\lim }\limits_{x \to 0} {{\sin x} \over x} = 1} \right\}

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