Question 351 of 401MediumMCQ
The value of
\[\mathop {\lim }\limits_{x \to 0} \left( {{x \over {\root 8 \of {1 - \sin x} - \root 8 \of {1 + \sin x} }}} \right)\] is equal to :
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Official solution
\[\mathop {\lim }\limits_{x \to 0} \left( {{x \over {\root 8 \of {1 - \sin x} - \root 8 \of {1 + \sin x} }}} \right)\]
= \[\mathop {\lim }\limits_{x \to 0} {x \over {\left[ {{{\left( {1 - \sin x} \right)}^{{1 \over 8}}} - {{\left( {1 + \sin x} \right)}^{{1 \over 8}}}} \right]}} \times \left[ {{{{{\left( {1 - \sin x} \right)}^{{1 \over 8}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 8}}}} \over {{{\left( {1 - \sin x} \right)}^{{1 \over 8}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 8}}}}}} \right]\]
= \[\mathop {\lim }\limits_{x \to 0} {{x\left[ {{{\left( {1 - \sin x} \right)}^{{1 \over 8}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 8}}}} \right]} \over {\left[ {{{\left( {1 - \sin x} \right)}^{{1 \over 4}}} - {{\left( {1 + \sin x} \right)}^{{1 \over 4}}}} \right]}} \times \left[ {{{{{\left( {1 - \sin x} \right)}^{{1 \over 4}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 4}}}} \over {{{\left( {1 - \sin x} \right)}^{{1 \over 4}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 4}}}}}} \right]\]
= \[\mathop {\lim }\limits_{x \to 0} {{x\left[ 2 \right]\left[ 2 \right]} \over {\left[ {{{\left( {1 - \sin x} \right)}^{{1 \over 2}}} - {{\left( {1 + \sin x} \right)}^{{1 \over 2}}}} \right]}} \times \left[ {{{{{\left( {1 - \sin x} \right)}^{{1 \over 2}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 2}}}} \over {{{\left( {1 - \sin x} \right)}^{{1 \over 2}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 2}}}}}} \right]\]
= \[\mathop {\lim }\limits_{x \to 0} {{x\left[ 2 \right]\left[ 2 \right]\left[ 2 \right]} \over {\left[ {\left( {1 - \sin x} \right) - \left( {1 + \sin x} \right)} \right]}}\]
\[ = \mathop {\lim }\limits_{x \to 0} \left( { - {1 \over 2}} \right)(2)(2)(2)\]
= -4
\[\because\] \[\left\{ {\mathop {\lim }\limits_{x \to 0} {{\sin x} \over x} = 1} \right\}\]
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