To evaluate

$\lim_{x \to 1^+} \left( \ln(x) \tan\left(\frac{\pi x}{2}\right)\right),$

let us proceed step by step:

Step 1: Behavior of $\ln(x)$ as $x \to 1^+$

The natural logarithm $\ln(x)$ approaches $0$ as $x \to 1^+$. Hence, $\ln(x) \to 0$.

Step 2: Behavior of $\tan\left(\frac{\pi x}{2}\right)$ as $x \to 1^+$

As $x \to 1^+$, $\frac{\pi x}{2} \to \frac{\pi}{2}$. The tangent function $\tan(\theta)$ approaches $+\infty$ as $\theta \to \frac{\pi}{2}^-$.

Step 3: Combination of the terms

The limit involves the product of two terms: $\ln(x) \to 0 \quad \text{and} \quad \tan\left(\frac{\pi x}{2}\right) \to +\infty.$ This creates an indeterminate form of the type $0 \cdot \infty$.

Step 4: Rewrite the expression

Rewrite the product as: $\ln(x) \tan\left(\frac{\pi x}{2}\right) = \frac{\ln(x)}{\cot\left(\frac{\pi x}{2}\right)},$ where $\cot(y) = 1 / \tan(y)$.

Now, as $x \to 1^+$:

• $\ln(x) \to 0$,
• $\cot\left(\frac{\pi x}{2}\right) \to 0^+$.

Thus, the limit takes the indeterminate form $\frac{0}{0}$.

Step 5: Apply L'Hôpital's Rule

Using L'Hôpital's Rule, we differentiate the numerator and the denominator with respect to $x$: $\lim_{x \to 1^+} \frac{\ln(x)}{\cot\left(\frac{\pi x}{2}\right)} = \lim_{x \to 1^+} \frac{\frac{d}{dx} \ln(x)}{\frac{d}{dx} \cot\left(\frac{\pi x}{2}\right)}.$

Differentiate the numerator:

$\frac{d}{dx} \ln(x) = \frac{1}{x}.$

Differentiate the denominator:

$\frac{d}{dx} \cot\left(\frac{\pi x}{2}\right) = -\csc^2\left(\frac{\pi x}{2}\right) \cdot \frac{\pi}{2}.$

Substitute these derivatives into the limit: $\lim_{x \to 1^+} \frac{\ln(x)}{\cot\left(\frac{\pi x}{2}\right)} = \lim_{x \to 1^+} \frac{\frac{1}{x}}{-\csc^2\left(\frac{\pi x}{2}\right) \cdot \frac{\pi}{2}}.$

Simplify:

$= \lim_{x \to 1^+} \frac{-2}{x \pi \csc^2\left(\frac{\pi x}{2}\right)}.$

As $x \to 1^+$, $\csc^2\left(\frac{\pi x}{2}\right) \to \csc^2\left(\frac{\pi}{2}\right) = 1$, and $x \to 1$. Thus, the limit becomes: $\lim_{x \to 1^+} \frac{-2}{\pi} = -\frac{2}{\pi}.$

Final Answer:

$\lim_{x \to 1^+} \left( \ln(x) \tan\left(\frac{\pi x}{2}\right) \right) = -\frac{2}{\pi}.$

Would you like to see the process for another example or related explanation?

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Related Questions:

1. What is the significance of L'Hôpital's Rule in solving limits?
2. How do trigonometric limits help in analyzing asymptotic behavior?
3. Can you explain limits of logarithmic functions in other contexts?
4. Why does $\ln(x)$ go to zero as $x \to 1^+$?
5. What are other methods to analyze indeterminate forms like $0 \cdot \infty$?

Tip: When working with limits involving logarithmic and trigonometric functions, rewriting the product as a quotient often simplifies the evaluation process!