Limit Evaluation: \(\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{x\sin x}}\)

To determine the value of the given limit, we begin by substituting \(x = 0\) into the expression to observe its form.

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

Since the limit results in the indeterminate form \(\frac{0}{0}\), we can proceed by either applying L'Hopital's Rule or by utilizing standard trigonometric limit identities.

Method 1: Applying L'Hopital's Rule for Limit Calculation

L'Hopital's Rule is a powerful technique for evaluating limits of indeterminate forms. It states that if \(\mathop {\lim }\limits_{x \to c} \frac{{f(x)}}{{g(x)}}\) is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then we can find the limit by taking the derivatives of the numerator and denominator: \(\mathop {\lim }\limits_{x \to c} \frac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to c} \frac{{f'(x)}}{{g'(x)}}\), provided the latter limit exists.

Let's define our functions: \(f(x) = 1 - \cos x\) and \(g(x) = x\sin x\).

Step 1: Calculate the first derivatives of \(f(x)\) and \(g(x)\).

• Derivative of \(f(x)\): \(f'(x) = \frac{d}{dx}(1 - \cos x) = 0 - (-\sin x) = \sin x\)
• Derivative of \(g(x)\): \(g'(x) = \frac{d}{dx}(x\sin x)\) (This requires the product rule \((uv)' = u'v + uv'\))
  • Let \(u = x\) and \(v = \sin x\).
  • Then \(u' = 1\) and \(v' = \cos x\).
  • So, \(g'(x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x\cos x\).

The limit expression now becomes: \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{\sin x + x\cos x}}\)

Step 2: Re-check the form of the limit.

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

The limit is still in the \(\frac{0}{0}\) indeterminate form, which means we need to apply L'Hopital's Rule one more time.

Step 3: Calculate the second derivatives.

• Second derivative of \(f(x)\): \(f''(x) = \frac{d}{dx}(\sin x) = \cos x\)
• Second derivative of \(g(x)\): \(g''(x) = \frac{d}{dx}(\sin x + x\cos x)\)
  • Derivative of \(\sin x\) is \(\cos x\).
  • Derivative of \(x\cos x\) (using product rule again) is \(1 \cdot \cos x + x \cdot (-\sin x) = \cos x - x\sin x\).
  • Therefore, \(g''(x) = \cos x + \cos x - x\sin x = 2\cos x - x\sin x\).

The limit expression transforms into: \(\mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{2\cos x - x\sin x}}\)

Step 4: Substitute \(x = 0\) into the latest expression.

\(\frac{{\cos(0)}}{{2\cos(0) - 0\sin(0)}} = \frac{{1}}{{2 \cdot 1 - 0 \cdot 0}} = \frac{{1}}{{2 - 0}} = \frac{{1}}{{2}}\)

Method 2: Using Standard Limit Identities for Limit Calculation

This approach simplifies the problem by leveraging well-known trigonometric limit formulas.

We recall two essential standard limits as \(x \to 0\):

• \(\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{x^2}} = \frac{1}{2}\)
• \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1\)

Let's rewrite the given expression: \(\frac{{1 - \cos x}}{{x\sin x}}\)

To align with our standard limits, we can divide both the numerator and the denominator by \(x^2\):

\[ \frac{{1 - \cos x}}{{x\sin x}} = \frac{{\frac{{1 - \cos x}}{{x^2}}}}{{\frac{{x\sin x}}{{x^2}}}} = \frac{{\frac{{1 - \cos x}}{{x^2}}}}{{\frac{{\sin x}}{x}}} \]

Now, we can apply the limit to the modified expression. Since the limit of a quotient is the quotient of the limits (provided the denominator's limit is not zero):

\[ \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{1 - \cos x}}{{x^2}}}}{{\frac{{\sin x}}{x}}} = \frac{{\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{x^2}}}}{{\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}}} \]

Substitute the known values of the standard limits:

\[ = \frac{{1/2}}{{1}} = \frac{{1}}{{2}} \]

Conclusion on Limit Value

Both methods, L'Hopital's Rule and using standard limit identities, consistently show that the value of the limit \(\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{x\sin x}}\) is \(\frac{1}{2}\).