Problem 24 - Calculus - AP Edition

Question

Find the limits. $$\lim _{x \rightarrow 0^{+}} \tan x \ln x$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Analyze the Given Limit

We are asked to find the limit \( \lim _{x \rightarrow 0^{+}} \tan x \ln x \). The limit involves a product of functions, \( \tan x \) and \( \ln x \), both of which have behavior that needs to be scrutinized as \( x \to 0^{+} \).

Step 2: Examine \( \tan x \) as \( x \to 0^{+} \)

The tangent function \( \tan x \) approaches \( 0 \) as \( x \to 0^{+} \) because \( \tan 0 = 0 \). This tells us one component of the product approaches \( 0 \).

Step 3: Examine \( \ln x \) as \( x \to 0^{+} \)

The natural logarithm function \( \ln x \) approaches \( -\infty \) as \( x \to 0^{+} \). This is because the logarithm function goes to \( -\infty \) as its argument approaches 0 from the positive side.

Step 4: Determine the Indeterminate Form

The product \( \tan x \ln x \) as \( x \rightarrow 0^{+} \) becomes \( 0 \times (-\infty) \), which is an indeterminate form of type \( 0 \times \infty \). This suggests using L'Hôpital's Rule might be a good idea, but it requires a \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form.

Step 5: Apply L'Hôpital's Rule with a Transformation

To use L'Hôpital's Rule, express the limit in a fraction form:\[\lim_{x \to 0^+} \tan x \ln x = \lim_{x \to 0^+} \frac{\ln x}{\cot x}.\]Here, we transform \( \tan x \) into \( \cot x = \frac{1}{\tan x} \), so notice \( \cot x \to \infty \) as \( x \to 0^+ \). This limit is now of the form \( \frac{-\infty}{\infty} \).

Step 6: Differentiate Numerator and Denominator

Apply L'Hôpital's Rule:Differentiate \( \ln x \) to get \( \frac{1}{x} \) and differentiate \( \cot x \) to get \( -\csc^2 x \). The limit becomes:\[\lim_{x \to 0^+} \frac{\frac{d}{dx}[\ln x]}{\frac{d}{dx}[\cot x]} = \lim_{x \to 0^+} \frac{1/x}{-\csc^2 x} = \lim_{x \to 0^+} -\frac{x}{\sin^2 x}.\]

Step 7: Evaluate the New Limit

Now consider \( \lim_{x \to 0^+} -\frac{x}{\sin^2 x} \). As \( x \to 0^+ \), \( \sin x \to x \), thus \( \sin^2 x \to x^2 \). So,\[\lim_{x \to 0^+} -\frac{x}{\sin^2 x} = \lim_{x \to 0^+} -\frac{x}{x^2} = \lim_{x \to 0^+} -\frac{1}{x} = 0.\]

Step 8: Conclusion

The limit \( \lim _{x \rightarrow 0^{+}} \tan x \ln x = 0 \).

Key Concepts

Indeterminate Forms

In calculus, we often encounter expressions that do not have an obvious limit as they approach certain values. These are known as indeterminate forms. The typical types include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), and the product type \(0 \times \infty\). In our exercise, we tackled the expression \(\tan x \ln x\) as \(x \rightarrow 0^+\). Here, \(\tan x\) tends towards 0 and \(\ln x\) diverges to \(-\infty\). This yields an indeterminate form of the type \(0 \times (-\infty)\).

Understanding indeterminate forms is crucial because they signal that direct substitution in the limit does not provide an answer. Instead, we need to apply algebraic transformations or calculus techniques, such as L'Hôpital's Rule, to resolve the indeterminacy. Recognizing and resolving these forms is essential for limit evaluation.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to find limits of indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Remarkably, it allows the transformation of a complex limit problem into one solvable by derivatives.

To apply L'Hôpital's Rule, one must transform the problematic expression into a rational form that fits the rule's conditions. In our exercise, we initially encountered \(0 \times (-\infty)\) and transformed it into the rational form \(\frac{-\infty}{\infty}\) by rewriting \(\tan x \ln x\) as \(\frac{\ln x}{\cot x}\). This setup allowed the application of the rule.

We differentiated the numerator and denominator: the derivative of \(\ln x\) is \(\frac{1}{x}\) and the derivative of \(\cot x\) is \(-\csc^2 x\). Thus, the new limit expression \(-\frac{x}{\sin^2 x}\) becomes straightforward to evaluate. Without L'Hôpital's Rule, conquering such limits would be significantly more challenging.

Trigonometric Functions

Trigonometric functions, like sine, cosine, and tangent, are core components in many calculus problems, and understanding their behavior is crucial for solving limits. In this problem, the tangent function \(\tan x\) plays a significant role. As \(x\) approaches 0 from the positive side, we observe that \(\tan x\) approaches 0.

This understanding helps identify the initial indeterminate form when combined with the logarithmic function. Furthermore, when transforming \(\tan x\) into \(\cot x = \frac{1}{\tan x}\), it was key to know that \(\cot x\) would tend towards infinity as \(x\) gets closer to 0.

Recognizing these behaviors can dramatically simplify the evaluation of limits involving trigonometric functions, especially when they are part of more complex products or quotients.