To evaluate the given limit:

$\lim_{x \to 1} (1 - x) \tan \frac{\pi x}{2},$

we proceed as follows:

Step 1: Analyze the behavior of $(1 - x)$ and $\tan \frac{\pi x}{2}$ as $x \to 1$

• As $x \to 1$, $1 - x \to 0$.
• Simultaneously, $\frac{\pi x}{2} \to \frac{\pi}{2}$, and thus $\tan \frac{\pi x}{2} \to \infty$ because the tangent function diverges near $\frac{\pi}{2}$.

Thus, we are evaluating the product of a term tending to $0$ and another term tending to $\infty$, which is an indeterminate form.

Step 2: Substitute $x = 1 - h$ (with $h \to 0^+$)

Let $x = 1 - h$, where $h \to 0$ as $x \to 1$. Substituting in the original expression:

$\lim_{x \to 1} (1 - x) \tan \frac{\pi x}{2} = \lim_{h \to 0} h \tan \frac{\pi (1 - h)}{2}.$

Simplify the argument of the tangent:

$\frac{\pi (1 - h)}{2} = \frac{\pi}{2} - \frac{\pi h}{2}.$

Thus, the expression becomes:

$\lim_{h \to 0} h \tan \left( \frac{\pi}{2} - \frac{\pi h}{2} \right).$

Step 3: Use the identity for $\tan \left( \frac{\pi}{2} - \theta \right)$

From trigonometric identities:

$\tan \left( \frac{\pi}{2} - \theta \right) = \cot \theta.$

Here, $\theta = \frac{\pi h}{2}$, so:

$\tan \left( \frac{\pi}{2} - \frac{\pi h}{2} \right) = \cot \frac{\pi h}{2}.$

Therefore, the limit becomes:

$\lim_{h \to 0} h \cot \frac{\pi h}{2}.$

Step 4: Approximate $\cot \frac{\pi h}{2}$ for small $h$

For small values of $\theta$, $\cot \theta \approx \frac{1}{\theta}$. Here, $\theta = \frac{\pi h}{2}$, so:

$\cot \frac{\pi h}{2} \approx \frac{1}{\frac{\pi h}{2}} = \frac{2}{\pi h}.$

Substituting this approximation into the limit:

$\lim_{h \to 0} h \cot \frac{\pi h}{2} \approx \lim_{h \to 0} h \cdot \frac{2}{\pi h}.$

Simplify:

$\lim_{h \to 0} \frac{2h}{\pi h} = \frac{2}{\pi}.$

Final Answer:

The value of the limit is:

$\boxed{\frac{2}{\pi}}$

Let me know if you need further clarification or step-by-step details!

Related Questions:

1. What is the limit of $\lim_{x \to 0} x \tan \frac{1}{x}$?
2. Evaluate $\lim_{x \to 0} \frac{\sin x}{x}$ and explain its significance.
3. Analyze $\lim_{x \to \frac{\pi}{2}} (\frac{\pi}{2} - x) \tan x$.
4. How can you use Taylor series to approximate $\tan x$ near $x = 0$?
5. Explain why $\tan \left( \frac{\pi}{2} - \theta \right) = \cot \theta$ using the unit circle.

Tip:

When dealing with limits involving trigonometric functions near singularities (like $\frac{\pi}{2}$), substitution and small-angle approximations are powerful tools!