Question

Use series to evaluate the limit.

\(\mathop {lim}\limits_{x \to 0} \frac{{sinx - x + \frac{1}{6}{x^3}}}{{{x^5}}}\)

Short answer

Therefore, the value of the given limit is: \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - x + \frac{1}{6}{x^3}}}{{{x^5}}} = \frac{1}{{120}}\)

Step-by-step solution

Step 1: Given information

The given limit is:

\(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - x + \frac{1}{6}{x^3}}}{{{x^5}}}\)

Note that the series for \(\sin x\)is given by:

\(\sin x = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{x^{(2n + 1)}}}}{{(2n + 1)!}} = x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - \frac{{{x^7}}}{{7!}} + \frac{{{x^9}}}{{9!}} - \ldots \)

Step 2: Substitution the expression into the limits

\(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - x + \frac{1}{6}{x^3}}}{{{x^5}}}\)

Substitute\(\sin x = x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - \frac{{{x^7}}}{{7!}} + \frac{{{x^9}}}{{9!}} - \ldots \)into the limit expression:\(\mathop {\lim }\limits_{x \to 0} \frac{{x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - \frac{{{x^7}}}{{7!}} + \frac{{{x^9}}}{{9!}} - \ldots - x + \frac{1}{6}{x^3}}}{{{x^5}}}\)

Substitute \(3! = 6,5! = 120,7! = 5040\) and \(9! = 362880\) into the expression:

\(\mathop {\lim }\limits_{x \to 0} \frac{{x - \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{120}} - \frac{{{x^7}}}{{5040}} + \frac{{{x^9}}}{{362880}} - \ldots - x + \frac{1}{6}{x^3}}}{{{x^5}}}\)

Step 3: Simplify the expression

\(\mathop {\lim }\limits_{x \to 0} \frac{{0 + 0 + \frac{{{x^5}}}{{120}} - \frac{{{x^7}}}{{5040}} + \frac{{{x^9}}}{{362880}} - \ldots }}{{{x^5}}}\)

Rewrite the expression as:

\(\mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{x^5}}}{{120}} - \frac{{{x^7}}}{{5040}} + \frac{{{x^9}}}{{362880}} - \cdots }}{{{x^5}}}\)\(\mathop {\lim }\limits_{x \to 0} \frac{{{x^5}}}{{120 \cdot {x^5}}} - \frac{{{x^7}}}{{5040 \cdot {x^5}}} + \frac{{{x^9}}}{{362880 \cdot {x^5}}} - \ldots \)\(\mathop {\lim }\limits_{x \to 0} \frac{{\left( {\frac{1}{{2!}} - \frac{{{x^2}}}{{4!}} + \frac{{{x^4}}}{{6!}} - ...} \right)}}{{ - \left( {\frac{1}{{2!}} + \frac{x}{{3!}} + \frac{{{x^2}}}{{4!}} + ..} \right)}}\)

Step 4: Substitute the terms

\(\begin{array}{l} = \frac{{{x^5}}}{{120{x^5}}}\\ = \frac{1}{{120}}\end{array}\)

\(\begin{array}{l} = \frac{{{x^7}}}{{5040{x^5}}}\\ = \frac{{{x^2}}}{{5040}}\end{array}\)

Simplify the terms:

\(\begin{array}{l}=\frac{{{x^9}}}{{362880{x^5}}}\\=\frac{{{x^4}}}{{362880}}\end{array}\)\(\mathop {\lim }\limits_{x \to 0} \frac{{{x^5}}}{{120 \cdot {x^5}}} - \frac{{{x^7}}}{{5040 \cdot {x^5}}} + \frac{{{x^9}}}{{362880 \cdot {x^5}}} - \ldots \)

Substitute \(\frac{{{x^5}}}{{120 \cdot {x^5}}} = \frac{1}{{120}}\)and \(\frac{{{x^7}}}{{5040 \cdot {x^5}}} = \frac{{{x^2}}}{{5040}}\) and \(\frac{{{x^9}}}{{362880 \cdot {x^5}}} = \frac{{{x^4}}}{{362880}}\)into the expression:\(\mathop {\lim }\limits_{x \to 0} \frac{1}{{120}} - \frac{{{x^2}}}{{5040}} + \frac{{{x^4}}}{{362880}} - \ldots \)

To determine the limit

To solve for the limit, substitute\(x = 0:\)

\( = \frac{1}{{120}} - \frac{{{0^2}}}{{5040}} + \frac{{{0^4}}}{{362880}} \ldots \)

Note that when the fraction's numerator is\(0,\)the fraction will also be equal to\(0:\)

\( = \frac{1}{{120}} - 0 + 0 \ldots \)\( = \frac{1}{{120}}\)

Evaluate the limit

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin (x) - x + \frac{1}{6}{x^3}}}{{{x^5}}}} \right)\)

Use the L'Hopital's rule

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\frac{d}{{dx}}\left( {\sin (x) - x + \frac{1}{6}{x^3}} \right)}}{{\frac{d}{{dx}}\left( {{x^5}} \right)}}} \right)\)\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\cos (x) - 1 + \frac{1}{2}{x^2}}}{{5{x^4}}}} \right)\)\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\frac{{2\cos (x) - 2 + {x^2}}}{2}}}{{5{x^4}}}} \right)\)\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{2\cos (x) - 2 + {x^2}}}{{10{x^4}}}} \right)\)

Use the L'Hopital's rule

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\frac{d}{{dx}}\left( {2\cos (x) - 2 + {x^2}} \right)}}{{\frac{d}{{dx}}\left( {10{x^4}} \right)}}} \right)\)\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{ - 2\sin (x) + 2x}}{{40{x^3}}}} \right)\)\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{ - \sin (x) + x}}{{20{x^3}}}} \right)\)\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\frac{d}{{dx}}( - \sin (x) + x)}}{{\frac{d}{{dx}}\left( {20{x^3}} \right)}}} \right)\)\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin (x)}}{{120x}}} \right)\)\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\frac{d}{{dx}}(\sin (x))}}{{\frac{d}{{dx}}(120x)}}} \right)\)\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\cos (x)}}{{120}}} \right)\)\(\frac{{\cos (0)}}{{120}}\)\( = \frac{1}{{120}}\)

Therefore, the value of the given limit is: \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - x + \frac{1}{6}{x^3}}}{{{x^5}}} = \frac{1}{{120}}\)