To find the value of the given limit, we need to evaluate the expression:

\[ \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{\sin x}}} \right) \]

Limit Evaluation Approach

First, we combine the two fractions into a single one. This is a crucial step for limits involving differences of fractions, as it often reveals the indeterminate form necessary for applying rules like L'Hôpital's.

Combine the terms:

\[ \frac{1}{x} - \frac{1}{\sin x} = \frac{\sin x - x}{x \sin x} \]

Now, let's substitute \(x=0\) into the combined expression to determine its form:

• Numerator: \(\sin(0) - 0 = 0 - 0 = 0\)
• Denominator: \(0 \cdot \sin(0) = 0 \cdot 0 = 0\)

Since we obtain the indeterminate form \(\frac{0}{0}\), we can proceed by applying L'Hôpital's Rule.

L'Hôpital's Rule Application

L'Hôpital's Rule is a powerful technique used to evaluate limits of indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). It states that if \(\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)}\) is an indeterminate form, then \(\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)} = \mathop {\lim }\limits_{x \to c} \frac{f'(x)}{g'(x)}\), provided the latter limit exists.

Let \(f(x) = \sin x - x\) and \(g(x) = x \sin x\).

Calculate the first derivatives of \(f(x)\) and \(g(x)\):

• The derivative of \(f(x)\) is \(f'(x) = \frac{d}{dx}(\sin x - x) = \cos x - 1\).
• For the derivative of \(g(x) = x \sin x\), we apply the product rule \((uv)' = u'v + uv'\). Let \(u = x\) and \(v = \sin x\). Then \(u' = 1\) and \(v' = \cos x\). Therefore, \(g'(x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x\).

So, the limit expression now becomes:

\[ \mathop {\lim }\limits_{x \to 0} \left( \frac{\cos x - 1}{\sin x + x \cos x} \right) \]

Let's substitute \(x=0\) again into this new expression to check its form:

• Numerator: \(\cos(0) - 1 = 1 - 1 = 0\)
• Denominator: \(\sin(0) + 0 \cdot \cos(0) = 0 + 0 \cdot 1 = 0\)

We still have the indeterminate form \(\frac{0}{0}\). This indicates that we need to apply L'Hôpital's Rule one more time.

Second L'Hôpital's Rule Application

We will now find the second derivatives of \(f(x)\) and \(g(x)\) to apply L'Hôpital's Rule for the second time.

• The second derivative of \(f(x)\) is \(f''(x) = \frac{d}{dx}(\cos x - 1) = -\sin x\).
• For the second derivative of \(g(x) = \sin x + x \cos x\), we differentiate term by term. The derivative of \(\sin x\) is \(\cos x\). For the term \(x \cos x\), we use the product rule again. Let \(u = x\) and \(v = \cos x\). Then \(u' = 1\) and \(v' = -\sin x\). So, \(\frac{d}{dx}(x \cos x) = 1 \cdot \cos x + x \cdot (-\sin x) = \cos x - x \sin x\). Combining these, \(g''(x) = \cos x + (\cos x - x \sin x) = 2 \cos x - x \sin x\).

Now, the limit expression transforms to:

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

Finally, substitute \(x=0\) into this expression to find the limit value:

• Numerator: \(-\sin(0) = -0 = 0\)
• Denominator: \(2 \cos(0) - 0 \cdot \sin(0) = 2 \cdot 1 - 0 \cdot 0 = 2 - 0 = 2\)

So, the value of the limit is:

\[ \frac{0}{2} = 0 \]

Final Limit Result

The value of the limit \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{\sin x}}} \right)\) is \(0\).

This detailed step-by-step approach using repeated applications of L'Hôpital's Rule is a standard method for evaluating such indeterminate limit forms in calculus.