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# Solve this limit problem

Note by Raja Metronetizen
4 years, 1 month 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

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

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

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

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

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

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago