Problem 22 - Thomas Calculus

Question

Use l'Hopital's rule to find the limits in Exercises \(7-50\) . $$ \lim _{x \rightarrow \pi / 2} \frac{\ln (\csc x)}{(x-(\pi / 2))^{2}} $$

Step-by-Step Solution

Answer

The limit diverges to \(-\infty\).

Step 1: Identify Indeterminate Form

First, we need to check if the limit has an indeterminate form. As \( x \to \frac{\pi}{2} \), \( \csc x = \frac{1}{\sin x} \) goes to infinity and \( \ln(\csc x) \to \ln(\infty) = \infty \). Meanwhile, \( (x - \frac{\pi}{2})^2 \to 0^+\). Therefore, the form is \( \frac{\infty}{0^+} \), which doesn't require l'Hôpital's Rule. There's a need to check if both the numerator and the denominator do indeed go to infinity as \( x \to \frac{\pi}{2} \). Proceed to check limits and derivatives.

Step 2: Simplify Expression

Rewrite \( \ln(\csc x) = \ln(\frac{1}{\sin x}) = -\ln(\sin x) \). Therefore, our limit becomes \[ \lim _{x \rightarrow \pi / 2} \frac{-\ln(\sin x)}{(x-(\pi / 2))^{2}}. \]

Step 3: Apply l'Hopital's Rule

Since both numerator and denominator approach infinity as \( x \to \frac{\pi}{2} \), we can apply l'Hopital's Rule, which involves differentiating the numerator and the denominator until the indeterminate form is resolved.

Step 4: Differentiate the Numerator and Denominator

Find the derivative of the numerator \( -\ln(\sin x) \): - Use chain rule: let \( u = \sin x \) then \(-\ln(u)' = -\frac{1}{u} \cdot u' = -\frac{1}{\sin x} \cdot \cos x = -\cot x \).Now find the derivative of the denominator \((x - \frac{\pi}{2})^2\):- Use power rule: derivative is \( 2(x - \frac{\pi}{2}) \).

Step 5: Compute the New Limit

Now compute the new limit: \[ \lim_{x \to \frac{\pi}{2}} \frac{-\cot x}{2(x - \frac{\pi}{2})} \]As \( x \to \frac{\pi}{2}, \cot x = \frac{\cos x}{\sin x} \to 0 \), but observe \(-\cot x\) and \(2(x - \frac{\pi}{2})\) tend towards \(\frac{0}{0}\), so l'Hopital's Rule might be applied again. Differentiate again.

Step 6: Apply l'Hopital's Rule Again

Differentiate \(-\cot x\) and \(2(x - \frac{\pi}{2})\) for a second application: - \(-\cot x\)'s derivative: \( \csc^2 x \). - \((2(x - \frac{\pi}{2}))'\)'s derivative: it is constant, \(2\).

Step 7: Evaluate the Second Differentiated Limit

Evaluate: \[ \lim_{x \to \frac{\pi}{2}} \frac{-\csc^2 x}{2} \]As \(x \to \frac{\pi}{2}\), \(-\csc^2 x \to -\infty\), indicating a limit that diverges.

Step 8: Conclude the Evaluation

In conclusion, we cannot find a finite limit for this expression as it diverges to negative infinity. Therefore, the limit is \(-\infty\).

Key Concepts

Indeterminate Forms

Indeterminate forms are expressions that don't initially provide clear information about limit behavior just by direct substitution. When solving a limits problem, especially involving functions like logarithms and trigonometric identities, it's common to encounter these forms. They often appear as ratios like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or other ambiguities such as \( \infty - \infty \).

In the given problem, as \( x \rightarrow \pi/2 \), \( \ln(\csc x) \) becomes divergent heading towards infinity, while \( (x - \pi/2)^2 \) tends to zero from the positive side. Hence, the form looks like \( \frac{\infty}{0^+} \), which doesn't match typical indeterminate categories but hints at a strong divergence. Nevertheless, resolving these kinds of expressions might sometimes lead us to undertakings with l'Hopital's Rule, provided the form suggests differentiable set-ups.

Differentiation

Differentiation is a powerful tool in calculus used to find the rate at which a function changes. It's essential in solving limit problems involving indeterminate forms. l'Hopital's Rule relies heavily on differentiation to tackle these forms by allowing us to differentiate the numerator and the denominator separately.

For the given exercise:

• The derivative of the numerator, \(-\ln(\sin x)\), is computed by employing the chain rule. We define a substitution \( u = \sin x \) which transforms the derivative to \(-\frac{1}{\sin x} \cdot \cos x = -\cot x\).
• For the denominator, \((x - \pi/2)^2\), we apply the power rule yielding \(2(x - \pi/2)\). This is quite straightforward, as it's merely a polynomial.

These derivatives assist in rewriting the limit expression thereby clarifying the form, eventually leading us to apply l'Hopital's Rule iteratively to gain more meaningful insights.

Trigonometric Limits

Trigonometric limits involve the behavior of trigonometric functions as they approach a particular point. They can often be tricky due to the oscillatory nature of these functions, but applying knowledge of limits with identities can greatly ease the process.

In the problem discussed, the term \( \csc x \), which equals \( \frac{1}{\sin x} \), becomes problematic as \( x \rightarrow \pi/2 \). Here, \( \sin x \rightarrow 1 \) leading \( \ln(\csc x) \) to head towards infinity due to the logarithm of a large quantity \( \ln(\infty) \). This is further coupled with \( \cot x = \frac{\cos x}{\sin x} \), which approaches zero as \( x \rightarrow \pi/2 \) because \( \cos(\pi/2) = 0 \).

Understanding these trigonometric behaviors supports the iterative application of l'Hopital's Rule by showing how trigonometric limits influence the differentiation steps and consequent evaluations of new limit forms that emerge. Ultimately, this clarifies why the original limit diverges to \(-\infty\).