CM

Calculus Level 4

$\large \lim_{\omega\to 0}\frac{\omega^{6000}-(\sin \omega)^{6000}}{\omega^{2}(\sin \omega)^{6000}} = \, ?$

6000 1000 1500 500

1 solution

Rishi Sharma
Jun 16, 2016

I will do the question for some generalised n instead of 6000.

Let $L=\lim _{ x\rightarrow 0 }{ \frac { { x }^{ n }-{ \left( \sin { x } \right) }^{ n } }{ { x }^{ 2 }{ \left( \sin { x } \right) }^{ n } } }$ Now as $x\rightarrow 0$ .We can assume $\sin{x} \approx x-\frac{{x}^{3}}{6}$ . Using Maclaurin Series of $\sin{x}$ .Substituting it in $L$ . We get $L=\lim _{ x\rightarrow 0 }{ \frac { { x }^{ n }-{ \left( x-\frac { { x }^{ 3 } }{ 6 } \right) }^{ n } }{ { x }^{ 2 }{ \left( x-\frac { { x }^{ 3 } }{ 6 } \right) }^{ n } } }$ Again using the argument that $x \rightarrow 0$ we can use binomial approximation and write ${ \left( x-\frac { { x }^{ 3 } }{ 6 } \right) }^{ n }={ x }^{ n }\left( 1-\frac { n{ x }^{ 2 } }{ 6 } \right)$ Substituting it in $L$ and doing some basic manipulations we get $L=\lim _{ x\rightarrow 0 }{ \frac { { x }^{ n+2 }\left( \frac { n }{ 6 } \right) }{ { x }^{ n+2 }\left( 1-\frac { n{ x }^{ 2 } }{ 6 } \right) } } =\frac { n }{ 6 }$ Now putting $n=6000$ we get $L=\frac { 6000 }{ 6 } =\boxed6$

Moderator note:

A much better (and more rigorous) way to express your solution is to use the Big-O notation directly. IE

$L=\lim _{ x\rightarrow 0 }{ \frac { { x }^{ n }-{ \left( \sin { x } \right) }^{ n } }{ { x }^{ 2 }{ \left( \sin { x } \right) }^{ n } } } =\lim _{ x\rightarrow 0 }{ \frac { { x }^{ n+2 }\left( \frac { n }{ 6 } \right) + O(x^{n+4}) }{ { x }^{ n+2 }\left( 1-\frac { n{ x }^{ 2 } }{ 6 } \right) + O(x^{n+3}) } } .$

This justifies why we can use the approximation of $\sin x$ and the binomial approximation.

Brilliant! . and thanks for generalisation :)

Prakhar Bindal - 4 years, 12 months ago

CM

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