Limit Evaluation Explained

Understanding limits is a fundamental concept in calculus. A limit describes the behavior of a function as its input approaches a certain value. In this problem, we need to find the value of the expression \(\mathop {\lim }\limits_{x \to 0} \frac{{{x^3} - {\rm{sin}}\left( x \right)}}{x}\).

Indeterminate Form Identification

First, let's substitute \(x = 0\) into the expression to see what form it takes:

• Numerator: \(0^3 - \sin(0) = 0 - 0 = 0\)
• Denominator: \(0\)

Since we get \(\frac{0}{0}\), this is an indeterminate form. When a limit results in an indeterminate form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can often use techniques like L'Hopital's Rule or series expansion to find the actual limit value.

Applying L'Hopital's Rule for Limit

L'Hopital's Rule states that if \(\mathop {\lim }\limits_{x \to c} \frac{{f\left( x \right)}}{{g\left( x \right)}}\) is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\mathop {\lim }\limits_{x \to c} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to c} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}\), provided the latter limit exists. In our case:

• Let \(f(x) = x^3 - \sin(x)\)
• Let \(g(x) = x\)

Now, let's find the derivatives of \(f(x)\) and \(g(x)\) with respect to \(x\):

• Derivative of the numerator, \(f'(x)\): \[ \frac{d}{dx}(x^3 - \sin(x)) = 3x^2 - \cos(x) \]
• Derivative of the denominator, \(g'(x)\): \[ \frac{d}{dx}(x) = 1 \]

Now, we can apply L'Hopital's Rule:

\[ \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} - {\rm{sin}}\left( x \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{3x^2 - {\rm{cos}}\left( x \right)}}{1} \]

Substitute \(x = 0\) into the new expression:

\[ = \frac{{3(0)^2 - {\rm{cos}}\left( 0 \right)}}{1} \]

Since \(\cos(0) = 1\):

\[ = \frac{{0 - 1}}{1} = -1 \]

Alternative Method: Series Expansion for Limit

We can also solve this limit using the Maclaurin series expansion for \(\sin(x)\) around \(x=0\). The series for \(\sin(x)\) is given by:

\[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots \]

Substitute this expansion into the original limit expression:

\[ \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} - {\rm{sin}}\left( x \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} - \left( {x - \frac{{x^3}}{{3!}} + \frac{{x^5}}{{5!}} - \dots } \right)}}{x} \]

Distribute the negative sign:

\[ = \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} - x + \frac{{x^3}}{{3!}} - \frac{{x^5}}{{5!}} + \dots }}{x} \]

Group terms and simplify:

\[ = \mathop {\lim }\limits_{x \to 0} \frac{{-x + x^3 + \frac{x^3}{6} - \frac{x^5}{120} + \dots }}{x} \]

Divide each term in the numerator by \(x\):

\[ = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{-x}}{x} + \frac{{x^3}}{x} + \frac{{x^3}}{6x} - \frac{{x^5}}{120x} + \dots } \right) \]

\[ = \mathop {\lim }\limits_{x \to 0} \left( {-1 + x^2 + \frac{x^2}{6} - \frac{x^4}{120} + \dots } \right) \]

Now, substitute \(x = 0\) into this simplified expression:

\[ = -1 + (0)^2 + \frac{(0)^2}{6} - \frac{(0)^4}{120} + \dots \]

\[ = -1 + 0 + 0 - 0 + \dots = -1 \]

Conclusion of Limit Value

Both L'Hopital's Rule and the series expansion method yield the same result. The value of the given limit is -1.