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Solve this limit problem

Please solve the limit problem and state the answer.

Note by Raja Metronetizen
3 years, 10 months ago

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Replace \(x\) with \( \sin (x) \), the expression becomes

\( \large \Rightarrow \displaystyle \lim_{x \to 0} \frac {x \space \sin ( \sin (x) ) - ( \sin^2 (x)) } { \sin^6 (x) } \)

The Maclaurin series of \( \sin (x) \) is \( x - \frac {1}{6} x^3 + \frac {1}{120} x^5 - \frac {1}{5040} x^7 + \ldots \)

For small \(x\), \( \sin (x) \approx x - \frac {1}{6} x^3 \)

\( \large \Rightarrow \displaystyle \lim_{x \to 0} \frac { x \space \sin ( x - \frac {1}{6} x^3 ) - (x - \frac {1}{6} x^3 )^2 }{ (x - \frac {1}{6} x^3)^6 } \)

\( \large \Rightarrow \displaystyle \lim_{x \to 0} \frac { x \space ( ( x - \frac {1}{6} x^3 ) - \frac {1}{6} ( x - \frac {1}{6} x^3 )^3 ) - (x - \frac {1}{6} x^3 )^2 }{ (x - \frac {1}{6} x^3)^6 } \)

After much simplifications

\( \large \Rightarrow \displaystyle \lim_{x \to 0} \frac {x^6 (x^4 - 18x^2 + 72)}{1296} \cdot \frac {1}{x^6 (1 - \frac {1}{6} x^2 )^6 } \)

\( \large \space \space \space \space = \displaystyle \lim_{x \to 0} \frac {x^4 - 18x^2 + 72}{1296} \cdot \frac {1}{ (1 - \frac {1}{6} x^2 )^6 } \)

\( \large \space \space \space \space = \frac {72}{1296} \cdot \frac {1}{1} = \frac {1}{18} \)


Alternatively, we can also apply the Maclaurin series of \( \sin (x) \) and \( \arcsin (x) \), but we need more terms

\( \sin (x) = x - \frac {1}{6} x^3 + \frac {1}{120} x^5 + O(x^7) \)

\( \arcsin (x) = x + \frac {1}{6} x^3 + \frac {3}{40} x^5 + O(x^7) \)

\( \large \Rightarrow \displaystyle \lim_{x \to 0} \frac { (x - \frac {1}{6} x^3 + \frac {1}{120} x^5 + O(x^7)) \cdot ( x + \frac {1}{6} x^3 + \frac {3}{40} x^5 + O(x^7) ) - x^2 } { x^6 } \)

\( \large \space \space \space \space = \displaystyle \lim_{x \to 0} \frac { x^6 ( \frac {3}{40} + \frac {1}{120} - \frac {1}{36} ) + O(x^8) } { x^6 } \)

\( \large \space \space \space \space = \displaystyle \lim_{x \to 0} ( \frac {3}{40} + \frac {1}{120} - \frac {1}{36} ) + O(x^2) \)

\( \large \space \space \space \space = \frac {3}{40} + \frac {1}{120} - \frac {1}{36} = \frac {1}{18} \) Pi Han Goh · 3 years, 10 months ago

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@Pi Han Goh thanks for details of the solution ...The answer is right...it's easy now to understand....thanks a lot... Raja Metronetizen · 3 years, 10 months ago

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