lim (x→1) (1-x) tan(πx/2) = .........

Q: Wait, why can't I just say $0 \times \infty = 0$?

That is the most common trap in calculus! Limits are about behavior, not static values. $0 \times \infty$ is an indeterminate form because the 'zero' and the 'infinity' are competing. One might shrink faster than the other grows, or vice versa. We must use algebra to reveal the true winner.

Q: Why did we choose to write $\tan \theta$ as $1/\cot \theta$ instead of $\sin \theta / \cos \theta$?

Great question! While $\sin/\cos$ is valid, it often leads to more complex derivatives. Using $\cot$ keeps the denominator as a single function, making the application of L'Hospital's Rule much cleaner and less prone to algebraic errors.

Q: Is there another way to solve this without L'Hospital's Rule?

Absolutely! You could use a substitution like $t = 1-x$. As $x \to 1$, $t \to 0$. Then $\tan(\frac{\pi(1-t)}{2}) = \tan(\frac{\pi}{2} - \frac{\pi t}{2}) = \cot(\frac{\pi t}{2})$. The limit becomes $t \cot(\frac{\pi t}{2})$, which is $\frac{t}{\tan(\pi t / 2)}$. Using the standard limit $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$, you get the same result. Both methods are elegant!

Question

$\lim_{x 	o 1} (1-x) 	an \frac{\pi x}{2} = .........$

Sigma Percentile · Accepted Answer

### Analyzing the Indeterminate Form

Imagine you are standing on the edge of a mathematical cliff, looking at the expression $\lim_{x 	o 1} (1-x) 	an \left( \frac{\pi x}{2} ight)$.

As you approach the value $x = 1$, you encounter a fascinating struggle. The term $(1-x)$ is shrinking toward zero, while $	an \left( \frac{\pi x}{2} ight)$ is racing toward infinity.

This is the classic $0 	imes \infty$ indeterminate form. It is a battle between a vanishing quantity and an exploding one, resulting in a specific, finite value waiting to be discovered.

### The Art of Transformation

We cannot simply multiply zero by infinity. We need to reshape this expression into a form where our calculus tools can work their magic.

L'Hospital's Rule is our most powerful weapon, but it demands a quotient: either $\frac{0}{0}$ or $\frac{\infty}{\infty}$. We can turn our product into a quotient by using the identity $	an 	heta = \frac{1}{\cot 	heta}$.

Rewriting the limit, we obtain:

$$\lim_{x 	o 1} \frac{1-x}{\cot \left( \frac{\pi x}{2} ight)}$$

If we test $x = 1$ again, the numerator is $1-1=0$, and the denominator is $\cot \left( \frac{\pi}{2} ight) = 0$. We have successfully created a $\frac{0}{0}$ form.

### The Calculus of Change

Now that we have a quotient, we apply L'Hospital's Rule, which states that the limit of the ratio of two functions is the same as the limit of the ratio of their derivatives.

The derivative of the numerator, $1-x$, is simply $-1$.

For the denominator, we differentiate $\cot \left( \frac{\pi x}{2} ight)$ using the chain rule. Since the derivative of $\cot u$ is $-\csc^2 u$, we multiply by the derivative of the inner function, $\frac{\pi}{2}$:

$$\frac{d}{dx} \left[ \cot \left( \frac{\pi x}{2} ight) ight] = -\csc^2 \left( \frac{\pi x}{2} ight) \cdot \frac{\pi}{2}$$

### The Final Victory

Let us assemble our new expression:

$$\lim_{x 	o 1} \frac{-1}{-\csc^2 \left( \frac{\pi x}{2} ight) \cdot \frac{\pi}{2}}$$

The negative signs cancel out, and the constant $\frac{\pi}{2}$ in the denominator flips to become $\frac{2}{\pi}$ in the numerator. We are left with:

$$\lim_{x 	o 1} \frac{2}{\pi \csc^2 \left( \frac{\pi x}{2} ight)}$$

Now, we can safely substitute $x = 1$. Since $\sin \left( \frac{\pi}{2} ight) = 1$, it follows that $\csc \left( \frac{\pi}{2} ight) = 1$. Squaring this value still yields $1$.

The entire expression simplifies beautifully to:

$$\frac{2}{\pi \cdot 1} = \frac{2}{\pi}$$

You have navigated the trap, performed the transformation, and executed the calculus. The hole in the graph at $x = 1$ is exactly at the height of **$\frac{2}{\pi}$**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: $lim_{x \to 1} (1 - x) tan \frac{π x}{2} = .........$

Visualized Solution

The Limit Problem

Direct Substitution

Identifying the Indeterminate Form

Strategy for $0 \times \infty$

Converting to a Fraction

Checking the New Form

Applying L'Hospital's Rule

Derivative of the Numerator

Derivative of the Denominator

The New Limit Expression

Simplifying the Expression

Evaluating the Limit

Calculating the Final Value

Final Answer

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Indeterminate Form

The Art of Transformation

The Calculus of Change

The Final Victory

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