New user? Sign up

Existing user? Sign in

Please solve the limit problem and state the answer.

Note by Raja Metronetizen 4 years, 5 months ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

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

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

Sort by:

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

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

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

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

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

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

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

Log in to reply

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

Problem Loading...

Note Loading...

Set Loading...

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

Sort by:

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

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

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

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

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

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

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

Log in to reply

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

Log in to reply