Key Concepts and Formulas

• Trigonometric Identities:
  • $1 - \cos 2x = 2 \sin^2 x$
• Standard Limits:
  • $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$
  • $\mathop {\lim }\limits_{y \to 0} \frac{\tan y}{y} = 1$ (which implies $\mathop {\lim }\limits_{y \to 0} \frac{y}{\tan y} = 1$)
• Limit Properties: The limit of a product is the product of the limits (if they exist), and the limit of a sum is the sum of the limits (if they exist).

Step-by-Step Solution

Step 1: Rewrite the expression and apply trigonometric identities. The given limit is: $L = \mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)} \over {x\tan 4x}}$ We can use the identity $1 - \cos 2x = 2 \sin^2 x$ to simplify the numerator. $L = \mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}x \left( {3 + \cos x} \right)} \over {x\tan 4x}}$

Step 2: Rearrange terms to utilize standard limits. To apply the standard limits $\frac{\sin x}{x} = 1$ and $\frac{\tan y}{y} = 1$, we need to manipulate the expression. We can separate the terms and multiply and divide by appropriate factors. $L = \mathop {\lim }\limits_{x \to 0} \left( \frac{{2{{\sin }^2}x}}{{{x^2}}} \cdot \frac{x}{{\tan 4x}} \cdot \left( {3 + \cos x} \right) \right)$ We've introduced $x^2$ in the denominator to match the $\sin^2 x$ term, and we have an extra $x$ in the numerator which we'll use with $\tan 4x$.

Step 3: Further rearrange for standard limits. We can rewrite the expression as: $L = \mathop {\lim }\limits_{x \to 0} \left( 2 \cdot \left(\frac{\sin x}{x}\right)^2 \cdot \frac{x}{{\tan 4x}} \cdot \left( {3 + \cos x} \right) \right)$ To use the $\frac{\tan y}{y} = 1$ limit, we need $\tan 4x$ to be divided by $4x$. We have $x$ in the numerator, so we'll multiply and divide by 4. $L = \mathop {\lim }\limits_{x \to 0} \left( 2 \cdot \left(\frac{\sin x}{x}\right)^2 \cdot \frac{4x}{{\tan 4x}} \cdot \frac{1}{4} \cdot \left( {3 + \cos x} \right) \right)$

Step 4: Apply limit properties and standard limits. Now we can apply the limit properties. We can take the constants out and evaluate the limits of the standard forms. $L = 2 \cdot \left(\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x}\right)^2 \cdot \left(\mathop {\lim }\limits_{x \to 0} \frac{4x}{{\tan 4x}}\right) \cdot \frac{1}{4} \cdot \mathop {\lim }\limits_{x \to 0} \left( {3 + \cos x} \right)$ We know that $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$. For $\mathop {\lim }\limits_{x \to 0} \frac{4x}{{\tan 4x}}$, let $y = 4x$. As $x \to 0$, $y \to 0$. So, $\mathop {\lim }\limits_{x \to 0} \frac{4x}{{\tan 4x}} = \mathop {\lim }\limits_{y \to 0} \frac{y}{{\tan y}} = 1$. And, $\mathop {\lim }\limits_{x \to 0} \left( {3 + \cos x} \right) = 3 + \cos 0 = 3 + 1 = 4$.

Step 5: Calculate the final value of the limit. Substitute the values of the individual limits back into the expression: $L = 2 \cdot (1)^2 \cdot (1) \cdot \frac{1}{4} \cdot 4$ $L = 2 \cdot 1 \cdot 1 \cdot \frac{1}{4} \cdot 4$ $L = 2 \cdot 1$ $L = 2$

Common Mistakes & Tips

• Incorrectly applying standard limits: Ensure that the argument of the trigonometric function matches the denominator. For example, for $\frac{\sin x}{x}$, the denominator must be $x$, not $2x$ or $4x$.
• Algebraic errors: Be careful with multiplying and dividing by constants. Introducing a factor of 4 in the denominator for $\tan 4x$ requires a corresponding factor of 4 in the numerator, or a factor of $\frac{1}{4}$ outside the limit.
• Forgetting the $1-\cos 2x$ identity: This identity is crucial for simplifying the expression and making it amenable to standard limit forms.

Summary

The problem requires evaluating a limit involving trigonometric functions. We begin by applying the double angle identity for cosine to simplify the numerator. Then, we strategically rearrange the expression to isolate terms that match the standard limit forms $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{y \to 0} \frac{y}{\tan y} = 1$. By carefully adjusting the terms with factors and applying the properties of limits, we evaluate each part of the expression separately and multiply the results to obtain the final limit value.

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