The value of lim (x → 1) [ (x² - 1) sin²(πx) ] / [ x⁴ - 2x³ + 2x - 1 ] is equal to:

Q: Why can't I just use L'Hôpital's Rule?

You absolutely could, but it would be a nightmare of product and chain rules! L'Hôpital's Rule is a powerful tool, but it often turns simple problems into algebraic messes. Factorization is cleaner, more elegant, and significantly less prone to calculation errors in this case.

Q: Why did the negative sign disappear when we squared $\sin(\pi + \pi h)$?

Great observation! Remember that $\sin(\pi + \theta) = -\sin(\theta)$. When we square this, we get $(-\sin(\theta))^2$. Since any negative number squared becomes positive, the negative sign vanishes. It's a beautiful property of the square function!

Q: How do I know when to use substitution?

Whenever you have a limit $x \to a$ and the expression involves $(x-a)$ or functions of $(x-a)$, substitution $h = x-a$ is your best friend. It shifts the limit to $h \to 0$, which allows you to use standard limit formulas like $\lim_{h \to 0} \frac{\sin h}{h} = 1$.

Question

The value of $\lim_{x 	o 1} \frac{(x^2 - 1) \sin^2(\pi x)}{x^4 - 2x^3 + 2x - 1}$ is equal to:

Options:

(A) $\frac{\pi^2}{6}$

(B) $\frac{\pi^2}{3}$

(C) $\frac{\pi^2}{2}$

(D) $\pi^2$

Sigma Percentile · Accepted Answer

### The Limit as a Journey

Welcome, fellow traveler of the mathematical landscape. Today, we are going to conquer a limit that, at first glance, might seem like a daunting wall of variables.

But remember, in the world of JEE Advanced, every complex expression is just a puzzle waiting for the right key. Our goal is to evaluate:

$$\lim_{x 	o 1} \frac{(x^2 - 1) \sin^2(\pi x)}{x^4 - 2x^3 + 2x - 1}$$

### Phase 1: The Indeterminate Trap

Before we dive into the deep end, we must always check the nature of our problem.

If we naively substitute $x = 1$, we find that the numerator becomes $(1^2 - 1) \sin^2(\pi) = 0 \cdot 0 = 0$, and the denominator becomes $1^4 - 2(1)^3 + 2(1) - 1 = 1 - 2 + 2 - 1 = 0$.

We are staring directly into the face of a $\frac{0}{0}$ indeterminate form. This is not a dead end; it is an invitation to simplify.

### Phase 2: Algebraic Surgery

Our primary suspect for the zero in the denominator is the polynomial $D = x^4 - 2x^3 + 2x - 1$. Since $x=1$ makes it zero, the Factor Theorem guarantees that $(x-1)$ is a factor.

Let's perform some surgical grouping to reveal it:

$$D = (x^4 - 1) - 2x^3 + 2x$$

$$D = (x^4 - 1) - 2x(x^2 - 1)$$

Using the difference of squares, $x^4 - 1 = (x^2 - 1)(x^2 + 1)$, we can rewrite this as:

$$D = (x^2 - 1)(x^2 + 1) - 2x(x^2 - 1)$$

Now, factor out the common $(x^2 - 1)$:

$$D = (x^2 - 1)(x^2 + 1 - 2x)$$

Look closely at that second bracket! It is the perfect square $(x - 1)^2$. Thus, our denominator is fully revealed as $D = (x^2 - 1)(x - 1)^2$.

### Phase 3: The Trigonometric Shift

With the denominator factored, our limit becomes:

$$\lim_{x 	o 1} \frac{(x^2 - 1) \sin^2(\pi x)}{(x^2 - 1)(x - 1)^2}$$

We can safely cancel the $(x^2 - 1)$ term because $x$ is approaching $1$, not equal to it. We are left with:

$$\lim_{x 	o 1} \frac{\sin^2(\pi x)}{(x - 1)^2}$$

To make this manageable, let's use the substitution $h = x - 1$. As $x 	o 1$, $h 	o 0$. Our expression transforms into:

$$\lim_{h 	o 0} \frac{\sin^2(\pi(1 + h))}{h^2} = \lim_{h 	o 0} \frac{\sin^2(\pi + \pi h)}{h^2}$$

Using the identity $\sin(\pi + 	heta) = -\sin(	heta)$, we get $\sin(\pi + \pi h) = -\sin(\pi h)$. Squaring this, the negative sign disappears, leaving us with $\sin^2(\pi h)$.

### Phase 4: The Final Victory

We are left with the limit:

$$\lim_{h 	o 0} \frac{\sin^2(\pi h)}{h^2}$$

To use the standard limit $\lim_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$, we need the denominator to be $(\pi h)^2$. So, we multiply and divide by $\pi^2$:

$$\pi^2 \cdot \lim_{h 	o 0} \left( \frac{\sin(\pi h)}{\pi h} ight)^2 = \pi^2 \cdot (1)^2 = \pi^2$$

And there it is! Through careful factorization and a strategic substitution, we have arrived at the elegant result of **$\pi^2$**. You have successfully navigated the complexity.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: The value of $lim_{x \to 1} \frac{( x ^{2} - 1 ) s i n ^{2} ( π x )}{x ^{4} - 2 x ^{3} + 2 x - 1}$ is equal to:

Select Answer:

Visualized Solution

Checking the Limit Form

Analyzing the Denominator

Grouping Terms for Factorization

Extracting Common Factors

Final Factorization of Denominator

Canceling Common Factors

Substitution for Simplification

Transforming the Numerator

Using Trigonometric Identities

Setting up the Standard Limit

Final Evaluation

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

The Limit as a Journey

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of limx→1​x4−2x3+2x−1(x2−1)sin2(πx)​ is equal to: