L'Hospital's Rule? (3)

Calculus Level 4

L = lim ⁡ x → 0 x ( a 2 − ( a − x ) 2 ) 1 3 ( ( 8 a x − 4 x 2 ) 1 3 + ( 8 a x ) 1 3 ) 4 \large L = \lim_{x\to0} \dfrac{x\left(a^2-(a-x)^2 \right)^{\frac13}}{\left( (8ax-4x^2)^{\frac13}+(8ax)^{\frac13} \right)^4}

Find the value of L L , where a a is a constant.

Limit does not exist 2 21 / 3 a 2^{{21} / 3}a a 2 11 / 3 \dfrac{a}{2^{{11} / 3}} 1 2 23 / 3 a \dfrac{1}{2^{{23} / 3}a} None of these choices

Sabhrant Sachan by Sabhrant Sachan , 21, India

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1 solution

Sabhrant Sachan
May 30, 2016

Relevant wiki: Limits of functions - Problem solving

L = lim ⁡ x → 0 x ( a 2 − ( a 2 + x 2 − 2 a x ) 2 ) 1 3 x 4 3 ( ( 8 a − 4 x ) 1 3 + ( 8 a ) 1 3 ) 4 L = lim ⁡ x → 0 ( 2 a x − x 2 ) 1 3 x 1 3 ( ( 8 a − 4 x ) 1 3 + ( 8 a ) 1 3 ) 4 L = lim ⁡ x → 0 x 1 3 ( 2 a − x ) 1 3 x 1 3 ( ( 8 a − 4 x ) 1 3 + ( 8 a ) 1 3 ) 4 L = 2 1 3 a 1 3 ( 2 ⋅ ( 8 a ) 1 3 ) 4 L = 1 2 23 2 a L = \lim_{x\to0} \dfrac{x\left(a^2-(a^2+x^2-2ax)^2 \right)^{\frac13}}{x^{\frac43}\left( (8a-4x)^{\frac13}+(8a)^{\frac13} \right)^4} \\ L = \lim_{x\to0} \dfrac{(2ax-x^2)^{\frac13}}{x^{\frac13}\left( (8a-4x)^{\frac13}+(8a)^{\frac13} \right)^4} \\ L = \lim_{x\to0} \dfrac{x^{\frac13}(2a-x)^{\frac13}}{x^{\frac13}\left( (8a-4x)^{\frac13}+(8a)^{\frac13} \right)^4} \\ L = \dfrac{2^{\frac13}a^{\frac13} }{\left( 2\cdot(8a)^{\frac13} \right)^4 } \\ \boxed{L= \dfrac{1}{2^{\frac{23}{2}}a}}

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