Key Concepts and Formulas

1. Indeterminate Forms and L'Hôpital's Rule: When a limit results in the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, L'Hôpital's Rule states that $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.
2. Leibniz Integral Rule (Fundamental Theorem of Calculus, Part I): This rule is used to differentiate an integral whose limits of integration are functions of the variable of differentiation. If $F(x) = \int_{a(x)}^{b(x)} f(t) dt$, then $F'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$.
3. Standard Trigonometric Limits: The limit $\lim_{y \to 0} \frac{\sin y}{y} = 1$ is crucial for evaluating limits involving trigonometric functions.

Step-by-Step Solution

We are asked to evaluate the limit: $L = \mathop {\lim }\limits_{x \to 1} \left( {{{\int\limits_0^{{{\left( {x - 1} \right)}^2}} {t\cos \left( {{t^2}} \right)dt} } \over {\left( {x - 1} \right)\sin \left( {x - 1} \right)}}} \right)$

Step 1: Identify the Indeterminate Form We first evaluate the numerator and the denominator as $x \to 1$. For the numerator, as $x \to 1$, the upper limit of the integral $(x-1)^2 \to (1-1)^2 = 0$. Thus, the integral becomes $\int_0^0 t\cos(t^2) dt$, which is $0$ because the upper and lower limits are identical. For the denominator, as $x \to 1$, we have $(1-1)\sin(1-1) = 0 \cdot \sin(0) = 0 \cdot 0 = 0$. Since the limit is of the form $\frac{0}{0}$, we can apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule - Differentiate Numerator and Denominator Let $N(x) = \int_0^{(x-1)^2} t\cos(t^2) dt$ and $D(x) = (x-1)\sin(x-1)$. We need to find $N'(x)$ and $D'(x)$.

• Differentiating the Numerator $N(x)$: We use the Leibniz Integral Rule. Here, $f(t) = t\cos(t^2)$, the upper limit is $b(x) = (x-1)^2$, and the lower limit is $a(x) = 0$. The derivative of the upper limit is $b'(x) = \frac{d}{dx}((x-1)^2) = 2(x-1)$. The derivative of the lower limit is $a'(x) = \frac{d}{dx}(0) = 0$. Applying Leibniz Rule: $N'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$ $N'(x) = \left( (x-1)^2 \cos(((x-1)^2)^2) \right) \cdot (2(x-1)) - \left( 0\cos(0^2) \right) \cdot 0$ $N'(x) = 2(x-1) \cdot (x-1)^2 \cos((x-1)^4)$ $N'(x) = 2(x-1)^3 \cos((x-1)^4)$

• Differentiating the Denominator $D(x)$: We use the product rule for $D(x) = (x-1)\sin(x-1)$. Let $u = x-1$ and $v = \sin(x-1)$. Then $u' = 1$ and $v' = \cos(x-1) \cdot 1 = \cos(x-1)$. $D'(x) = u'v + uv' = (1)\sin(x-1) + (x-1)\cos(x-1)$ $D'(x) = \sin(x-1) + (x-1)\cos(x-1)$

Now, applying L'Hôpital's Rule, the limit becomes: $L = \mathop {\lim }\limits_{x \to 1} \frac{2(x-1)^3 \cos((x-1)^4)}{\sin(x-1) + (x-1)\cos(x-1)}$

Step 3: Simplify and Re-evaluate the Limit Let's substitute $x=1$ into the new expression to check the form: Numerator: $2(1-1)^3 \cos((1-1)^4) = 2(0)^3 \cos(0) = 0$. Denominator: $\sin(1-1) + (1-1)\cos(1-1) = \sin(0) + 0 \cdot \cos(0) = 0 + 0 = 0$. The limit is still of the form $\frac{0}{0}$. To simplify, let $y = x-1$. As $x \to 1$, $y \to 0$. Substituting $y$ into the limit expression: $L = \mathop {\lim }\limits_{y \to 0} \frac{2y^3 \cos(y^4)}{\sin y + y\cos y}$ We can divide the numerator and denominator by $y$ (since $y \neq 0$ as $y \to 0$): $L = \mathop {\lim }\limits_{y \to 0} \frac{\frac{2y^3 \cos(y^4)}{y}}{\frac{\sin y + y\cos y}{y}} = \mathop {\lim }\limits_{y \to 0} \frac{2y^2 \cos(y^4)}{\frac{\sin y}{y} + \cos y}$

Step 4: Evaluate the Final Limit Now we can evaluate the limit by substituting $y=0$ and using the standard limit $\lim_{y \to 0} \frac{\sin y}{y} = 1$: Numerator: $2y^2 \cos(y^4) \to 2(0)^2 \cos(0^4) = 2 \cdot 0 \cdot \cos(0) = 0 \cdot 1 = 0$. Denominator: $\frac{\sin y}{y} + \cos y \to 1 + \cos(0) = 1 + 1 = 2$. Therefore, the limit is: $L = \frac{0}{2} = 0$

This corresponds to option (A).

Common Mistakes & Tips

1. Incorrectly Applying L'Hôpital's Rule: Ensure the limit is indeed in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying the rule.
2. Errors in Differentiation: Be meticulous when differentiating the numerator using the Leibniz rule and the denominator using the product rule. Small errors can lead to a completely different result.
3. Algebraic Simplification: After applying L'Hôpital's Rule, look for opportunities to simplify the expression using standard limits like $\lim_{y \to 0} \frac{\sin y}{y} = 1$ to avoid further complex differentiation.

Summary

The problem required evaluating a limit of an indeterminate form. We first identified the $\frac{0}{0}$ indeterminate form and applied L'Hôpital's Rule. This involved differentiating the numerator using the Leibniz Integral Rule and the denominator using the product rule. After the first application, the limit remained indeterminate. We then simplified the expression by substituting $y=x-1$ and dividing both numerator and denominator by $y$. Finally, we evaluated the simplified limit using standard trigonometric limits, yielding a result of $0$.

The final answer is $\boxed{0}$.