Limit Analysis

We are asked to evaluate the following limit:

$$ L = \mathop {\lim }\limits_{\theta \to \frac{\pi }{2}} \frac{{\log \left( {\theta - \frac{\pi }{2}} \right)}}{{\tan \theta }} $$

To begin, let's examine the behavior of the numerator and denominator as \(\theta\) approaches \(\frac{\pi}{2}\).

• Numerator: As \(\theta \to \frac{\pi}{2}\), the term \(\theta - \frac{\pi}{2}\) approaches 0. For the natural logarithm \(\log(y)\) to be defined in the set of real numbers, the argument \(y\) must be positive (\(y > 0\)). Therefore, we must consider the limit from the right side, where \(\theta > \frac{\pi}{2}\), meaning \(\theta - \frac{\pi}{2} \to 0^+\). In this case, the numerator \(\log \left( {\theta - \frac{\pi }{2}} \right)\) approaches \(-\infty\).
• Denominator: As \(\theta \to \frac{\pi}{2}\), the tangent function \(\tan \theta\) approaches infinity. Specifically, as \(\theta\) approaches \(\frac{\pi}{2}\) from the right (\(\theta \to (\frac{\pi}{2})^+\)), \(\tan \theta \to +\infty\).

Since the limit takes the indeterminate form \(\frac{-\infty}{+\infty}\), we can apply L'Hôpital's Rule.

L'Hôpital's Rule Application

L'Hôpital's Rule is a method for evaluating limits of fractions that result in indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). It states that the limit of the ratio of two functions is equal to the limit of the ratio of their derivatives, provided the latter limit exists.

Let the numerator function be \(f(\theta) = \log \left( {\theta - \frac{\pi }{2}} \right)\) and the denominator function be \(g(\theta) = \tan \theta\).

We need to find the derivatives of these functions:

• The derivative of the numerator is: $$ f'(\theta) = \frac{d}{d\theta} \left( \log \left( {\theta - \frac{\pi }{2}} \right) \right) = \frac{1}{\theta - \frac{\pi }{2}} $$
• The derivative of the denominator is: $$ g'(\theta) = \frac{d}{d\theta} (\tan \theta) = \sec^2 \theta $$

Now, we apply L'Hôpital's Rule:

$$ L = \mathop {\lim }\limits_{\theta \to \frac{\pi }{2}^+} \frac{f'(\theta)}{g'(\theta)} = \mathop {\lim }\limits_{\theta \to \frac{\pi }{2}^+} \frac{\frac{1}{\theta - \frac{\pi }{2}}}{\sec^2 \theta} $$

Simplifying the Derivative Ratio

Let's simplify the expression obtained after applying L'Hôpital's Rule:

$$ \frac{\frac{1}{\theta - \frac{\pi }{2}}}{\sec^2 \theta} = \frac{1}{\theta - \frac{\pi }{2}} \times \frac{1}{\sec^2 \theta} $$

Since \(\frac{1}{\sec^2 \theta} = \cos^2 \theta\), the expression becomes:

$$ \frac{\cos^2 \theta}{\theta - \frac{\pi }{2}} $$

To make the limit evaluation easier, let's use a substitution. Let \(x = \theta - \frac{\pi}{2}\). As \(\theta \to \frac{\pi}{2}^+\), we have \(x \to 0^+\). This also means \(\theta = x + \frac{\pi}{2}\).

Now substitute \(\theta = x + \frac{\pi}{2}\) into the expression:

• \(\cos \theta = \cos \left( x + \frac{\pi}{2} \right)\). Using the identity \(\cos(\frac{\pi}{2} + x) = -\sin x\).
• Therefore, \(\cos^2 \theta = (\cos \left( x + \frac{\pi}{2} \right))^2 = (-\sin x)^2 = \sin^2 x\).

The limit expression transforms into:

$$ L = \mathop {\lim }\limits_{x \to 0^+} \frac{\sin^2 x}{x} $$

Evaluating the Simplified Limit

We need to evaluate the limit \( \mathop {\lim }\limits_{x \to 0^+} \frac{\sin^2 x}{x} \).

We can rewrite this expression as a product:

$$ L = \mathop {\lim }\limits_{x \to 0^+} \left( \frac{\sin x}{x} \cdot \sin x \right) $$

By the product rule for limits, this is:

$$ L = \left( \mathop {\lim }\limits_{x \to 0^+} \frac{\sin x}{x} \right) \cdot \left( \mathop {\lim }\limits_{x \to 0^+} \sin x \right) $$

We use two well-known limits:

• The fundamental trigonometric limit: \(\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1\).
• The limit of \(\sin x\) as \(x \to 0\): \(\mathop {\lim }\limits_{x \to 0} \sin x = \sin(0) = 0\).

Substituting these values:

$$ L = 1 \cdot 0 $$

$$ L = 0 $$

Final Result

The evaluation shows that the limit of the given function \(\frac{{\log \left( {\theta - \frac{\pi }{2}} \right)}}{{\tan \theta }}\) as \(\theta\) approaches \(\frac{\pi}{2}\) is 0.