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Please solve the limit problem and state the answer.

Note by Raja Metronetizen 4 years, 1 month ago

Easy Math Editor

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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

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

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

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

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

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

\( \large \space \space \space \space = \frac {3}{40} + \frac {1}{120} - \frac {1}{36} = \frac {1}{18} \)

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thanks for details of the solution ...The answer is right...it's easy now to understand....thanks a lot...

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestReplace \(x\) 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

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

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

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

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

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

\( \large \space \space \space \space = \frac {3}{40} + \frac {1}{120} - \frac {1}{36} = \frac {1}{18} \)

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thanks for details of the solution ...The answer is right...it's easy now to understand....thanks a lot...

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