Key Concepts and Formulas

• Limit of a Rational Function: When evaluating limits, if direct substitution results in an indeterminate form (like $\frac{0}{0}$), we often need to simplify the expression by factoring or using algebraic manipulation.
• Factorization of Polynomials: Recognizing and applying polynomial factorization is crucial for simplifying rational functions. Specifically, the difference of squares formula ($a^2 - b^2 = (a-b)(a+b)$) and recognizing patterns like $(x-1)^2$ are useful.
• Standard Trigonometric Limit: The fundamental trigonometric limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ is essential for evaluating limits involving trigonometric functions. This can be extended to $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin^n \theta}{\theta^n} = 1$ for any positive integer $n$.
• Substitution in Limits: If we have a limit as $x \to a$, we can often perform a substitution $x = a+h$ (or $x = a-h$) to transform the limit into one as $h \to 0$. This is particularly helpful when dealing with terms like $(x-a)$.

Step-by-Step Solution

We are asked to evaluate the limit: $L = \mathop {\lim }\limits_{x \to 1} \frac{({x^2} - 1){\sin^2}(\pi x)}{{x^4} - 2{x^3} + 2x - 1}$

Step 1: Analyze the denominator. First, let's examine the denominator $D(x) = {x^4} - 2{x^3} + 2x - 1$. We notice that if we substitute $x=1$, we get $1^4 - 2(1)^3 + 2(1) - 1 = 1 - 2 + 2 - 1 = 0$. This indicates that $(x-1)$ is a factor of the denominator. Let's try to factor the denominator. We can rewrite the denominator as: $D(x) = {x^4} - 1 - 2{x^3} + 2x$ $D(x) = ({x^2} - 1)({x^2} + 1) - 2x({x^2} - 1)$ $D(x) = ({x^2} - 1)({x^2} + 1 - 2x)$ $D(x) = ({x^2} - 1)(x - 1)^2$

Step 2: Substitute the factored denominator back into the limit expression. Now, we replace the denominator with its factored form: $L = \mathop {\lim }\limits_{x \to 1} \frac{({x^2} - 1){\sin^2}(\pi x)}{({x^2} - 1){(x - 1)^2}}$

Step 3: Simplify the expression by canceling common factors. We observe that $(x^2 - 1)$ is a common factor in the numerator and the denominator. Since $x \to 1$, $x \neq 1$, so $x^2 - 1 \neq 0$. We can cancel this term: $L = \mathop {\lim }\limits_{x \to 1} \frac{{\sin^2}(\pi x)}{{(x - 1)^2}}$

Step 4: Perform a substitution to transform the limit to $h \to 0$. To handle the term $(x-1)$ in the denominator and the argument of the sine function, let's use the substitution $x = 1 + h$. As $x \to 1$, we have $1 + h \to 1$, which implies $h \to 0$. Substituting $x = 1 + h$ into the expression: $x - 1 = (1 + h) - 1 = h$ $\pi x = \pi (1 + h) = \pi + \pi h$

The limit expression becomes: $L = \mathop {\lim }\limits_{h \to 0} \frac{{\sin^2}(\pi + \pi h)}{{h^2}}$

Step 5: Use trigonometric identities. We know that $\sin(\pi + \theta) = -\sin(\theta)$. Therefore, $\sin(\pi + \pi h) = -\sin(\pi h)$. Squaring this, we get ${\sin^2}(\pi + \pi h) = (-\sin(\pi h))^2 = {\sin^2}(\pi h)$. So, the limit becomes: $L = \mathop {\lim }\limits_{h \to 0} \frac{{\sin^2}(\pi h)}{{h^2}}$

Step 6: Apply the standard trigonometric limit. We can rewrite the expression to fit the standard limit form $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. $L = \mathop {\lim }\limits_{h \to 0} \left( \frac{\sin(\pi h)}{h} \right)^2$ To use the standard limit, we need the denominator to be $\pi h$. We can multiply and divide by $\pi$: $L = \mathop {\lim }\limits_{h \to 0} \left( \frac{\sin(\pi h)}{\pi h} \times \pi \right)^2$ $L = \mathop {\lim }\limits_{h \to 0} \left( \pi \times \frac{\sin(\pi h)}{\pi h} \right)^2$ Using the property of limits that $\mathop {\lim }\limits_{y \to 0} (c \cdot f(y)) = c \cdot \mathop {\lim }\limits_{y \to 0} f(y)$ and $\mathop {\lim }\limits_{y \to 0} (f(y))^n = (\mathop {\lim }\limits_{y \to 0} f(y))^n$: $L = \pi^2 \times \left( \mathop {\lim }\limits_{h \to 0} \frac{\sin(\pi h)}{\pi h} \right)^2$ Let $\theta = \pi h$. As $h \to 0$, $\theta \to 0$. The limit inside the parenthesis is $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta}$, which is equal to 1. $L = \pi^2 \times (1)^2$ $L = \pi^2$

Common Mistakes & Tips

• Incorrect Factorization: Ensure the denominator is factored correctly. A common error is misidentifying the factors or making algebraic mistakes during factorization. Double-check by expanding the factored form.
• Handling the Sine Argument: When using the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, make sure the argument of the sine function and the denominator are identical. In this case, $\sin(\pi h)$ requires a denominator of $\pi h$, not just $h$.
• Indeterminate Form: Always check if direct substitution leads to an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$). If not, the limit can be found by direct substitution. In this problem, direct substitution of $x=1$ yields $\frac{0 \cdot \sin^2(\pi)}{0} = \frac{0}{0}$, confirming the need for simplification.

Summary

The problem involves evaluating a limit of a rational function multiplied by a squared trigonometric term. The key steps include factoring the denominator, simplifying the expression by canceling common factors, performing a substitution to transform the limit variable to approach zero, and finally applying the standard trigonometric limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. By carefully executing these steps, we arrive at the value of the limit.

The final answer is $\boxed{{\pi^2}}$.