Solve this limit problem

Please solve the limit problem and state the answer.

#HelpMe! #MathProblem #Math

Note by Sayan Chaudhuri
7 years, 8 months ago

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Comments

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

⇒lim⁡x→0x sin⁡(sin⁡(x))−(sin⁡2(x))sin⁡6(x) \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) \sin (x) is x−16x3+1120x5−15040x7+… x - \frac {1}{6} x^3 + \frac {1}{120} x^5 - \frac {1}{5040} x^7 + \ldots

For small xx, sin⁡(x)≈x−16x3 \sin (x) \approx x - \frac {1}{6} x^3

⇒lim⁡x→0x sin⁡(x−16x3)−(x−16x3)2(x−16x3)6 \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 }

⇒lim⁡x→0x ((x−16x3)−16(x−16x3)3)−(x−16x3)2(x−16x3)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

⇒lim⁡x→0x6(x4−18x2+72)1296⋅1x6(1−16x2)6 \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 }

    =lim⁡x→0x4−18x2+721296⋅1(1−16x2)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 }

    =721296⋅11=118 \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) \sin (x) and arcsin⁡(x) \arcsin (x) , but we need more terms

sin⁡(x)=x−16x3+1120x5+O(x7) \sin (x) = x - \frac {1}{6} x^3 + \frac {1}{120} x^5 + O(x^7)

arcsin⁡(x)=x+16x3+340x5+O(x7) \arcsin (x) = x + \frac {1}{6} x^3 + \frac {3}{40} x^5 + O(x^7)

⇒lim⁡x→0(x−16x3+1120x5+O(x7))⋅(x+16x3+340x5+O(x7))−x2x6 \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 }

    =lim⁡x→0x6(340+1120−136)+O(x8)x6 \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 }

    =lim⁡x→0(340+1120−136)+O(x2) \large \space \space \space \space = \displaystyle \lim_{x \to 0} ( \frac {3}{40} + \frac {1}{120} - \frac {1}{36} ) + O(x^2)

    =340+1120−136=118 \large \space \space \space \space = \frac {3}{40} + \frac {1}{120} - \frac {1}{36} = \frac {1}{18}

Pi Han Goh - 7 years, 8 months ago

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

Sayan Chaudhuri - 7 years, 8 months ago
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