Question

Use series to evaluate the following limit.

\(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - x}}{{{x^3}}}\)

Short answer

The required answer is \( - \frac{1}{6}\).

Step-by-step solution

Definition of Concept

Series : The sum of all the terms in a sequence can be used to generalise a series. However, there must be a clear relationship between all of the sequence's terms.

Step 2: Remembering the formula

Considering the formula,

\(\sin x = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{x^{2n + 1}}}}{{(2n + 1)!}} = x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - \frac{{{x^7}}}{{7!}} + \cdots \)

Simplifying the series

Putting the series for\({\rm{sinx}}\)and solving it,

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - x}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - \frac{{{x^7}}}{{7!}} + \cdots } \right) - x}}{{{x^3}}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{ - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - \frac{{{x^7}}}{{7!}} + \cdots }}{{{x^3}}}\\ = \mathop {\lim }\limits_{x \to 0} \left( { - \frac{1}{{3!}} + \frac{{{x^2}}}{{5!}} - \frac{{{x^4}}}{{7!}} + \cdots } \right)\\ = - \frac{1}{{3!}} + 0 - 0 + \cdots \\ = - \frac{1}{6}\end{array}\)

Therefore, the required solution is \({\rm{ - }}\frac{{\rm{1}}}{{\rm{6}}}\).