Key Concepts and Formulas

• Trigonometric Identities: The fundamental identity $1 - \cos^2(\theta) = \sin^2(\theta)$ will be used.
• Standard Limits:
  • $\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1$
  • $\lim_{\theta \to 0} \frac{\tan(\theta)}{\theta} = 1$
  • $\lim_{\theta \to 0} \frac{1 - \cos(\theta)}{\theta^2} = \frac{1}{2}$
  • $\lim_{\theta \to 0} \frac{\log_e(1+\theta)}{\theta} = 1$
• Limit Properties: The limit of a product is the product of the limits (provided they exist), and the limit of a quotient is the quotient of the limits (provided the denominator's limit is not zero).

Step-by-Step Solution

Let the given limit be $L$. $L = \lim_{x \rightarrow 0}\left(\left(\frac{1-\cos ^{2}(3 x)}{\cos ^{3}(4 x)}\right)\left(\frac{\sin ^{3}(4 x)}{\left(\log _{e}(2 x+1)\right)^{5}}\right)\right)$

Step 1: Simplify the trigonometric part using the identity $1 - \cos^2(\theta) = \sin^2(\theta)$. We observe the term $1 - \cos^2(3x)$ in the numerator of the first fraction. Applying the identity, we get $1 - \cos^2(3x) = \sin^2(3x)$. $L = \lim_{x \rightarrow 0}\left(\left(\frac{\sin ^{2}(3 x)}{\cos ^{3}(4 x)}\right)\left(\frac{\sin ^{3}(4 x)}{\left(\log _{e}(2 x+1)\right)^{5}}\right)\right)$

Step 2: Rearrange the terms to group related functions and prepare for standard limits. We want to use the standard limits $\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1$ and $\lim_{\theta \to 0} \frac{\log_e(1+\theta)}{\theta} = 1$. To do this, we need to introduce appropriate terms in the denominator. We can rewrite the expression as: $L = \lim_{x \rightarrow 0}\left(\frac{\sin ^{2}(3 x)}{(3x)^2} \cdot (3x)^2 \cdot \frac{1}{\cos ^{3}(4 x)} \cdot \frac{\sin ^{3}(4 x)}{(4x)^3} \cdot (4x)^3 \cdot \frac{1}{\left(\frac{\log _{e}(2 x+1)}{2 x} \cdot 2x\right)^{5}}\right)$ Let's rearrange this further to group terms that will approach 1: $L = \lim_{x \rightarrow 0}\left(\left(\frac{\sin ^{2}(3 x)}{(3x)^2}\right) \cdot \left(\frac{\sin ^{3}(4 x)}{(4x)^3}\right) \cdot \frac{1}{\left(\frac{\log _{e}(2 x+1)}{2 x}\right)^{5}} \cdot \frac{(3x)^2 \cdot (4x)^3}{\cos ^{3}(4 x) \cdot (2x)^5}\right)$

Step 3: Evaluate the limits of the standard forms. We have the following limits:

• $\lim_{x \rightarrow 0} \left(\frac{\sin(3x)}{3x}\right)^2 = \left(\lim_{x \rightarrow 0} \frac{\sin(3x)}{3x}\right)^2 = 1^2 = 1$.
• $\lim_{x \rightarrow 0} \left(\frac{\sin(4x)}{4x}\right)^3 = \left(\lim_{x \rightarrow 0} \frac{\sin(4x)}{4x}\right)^3 = 1^3 = 1$.
• $\lim_{x \rightarrow 0} \left(\frac{\log_e(2x+1)}{2x}\right)^5 = \left(\lim_{x \rightarrow 0} \frac{\log_e(2x+1)}{2x}\right)^5 = 1^5 = 1$.
• $\lim_{x \rightarrow 0} \cos(4x) = \cos(0) = 1$. Therefore, $\lim_{x \rightarrow 0} \cos^3(4x) = 1^3 = 1$.

Step 4: Simplify the remaining algebraic terms. Now we need to evaluate the limit of the remaining algebraic part: $\lim_{x \rightarrow 0} \frac{(3x)^2 \cdot (4x)^3}{\cos ^{3}(4 x) \cdot (2x)^5}$ Let's expand the powers: $(3x)^2 = 9x^2$ $(4x)^3 = 64x^3$ $(2x)^5 = 32x^5$ So the expression becomes: $\lim_{x \rightarrow 0} \frac{9x^2 \cdot 64x^3}{\cos ^{3}(4 x) \cdot 32x^5} = \lim_{x \rightarrow 0} \frac{9 \cdot 64 \cdot x^5}{32 \cdot x^5 \cdot \cos ^{3}(4 x)}$ We can cancel out $x^5$ from the numerator and denominator: $\lim_{x \rightarrow 0} \frac{9 \cdot 64}{32 \cdot \cos ^{3}(4 x)}$ Since $\lim_{x \rightarrow 0} \cos^3(4x) = 1$, the limit of this part is: $\frac{9 \cdot 64}{32 \cdot 1} = 9 \cdot \frac{64}{32} = 9 \cdot 2 = 18$

Step 5: Combine the limits of all parts. Now, we multiply the limits of all the grouped parts: $L = \left(\lim_{x \rightarrow 0}\left(\frac{\sin(3x)}{3x}\right)^2\right) \cdot \left(\lim_{x \rightarrow 0}\left(\frac{\sin(4x)}{4x}\right)^3\right) \cdot \left(\frac{1}{\lim_{x \rightarrow 0}\left(\frac{\log _{e}(2 x+1)}{2 x}\right)^{5}}\right) \cdot \left(\lim_{x \rightarrow 0} \frac{(3x)^2 \cdot (4x)^3}{\cos ^{3}(4 x) \cdot (2x)^5}\right)$ $L = (1) \cdot (1) \cdot \left(\frac{1}{1}\right) \cdot (18)$ $L = 1 \cdot 1 \cdot 1 \cdot 18 = 18$

Common Mistakes & Tips

• Incorrectly Applying Standard Limits: Ensure that the argument of the trigonometric function or logarithm matches the denominator exactly. For example, $\frac{\sin(3x)}{x}$ needs to be multiplied by 3 to become $\frac{\sin(3x)}{3x} \cdot 3$.
• Algebraic Errors: Be careful when simplifying powers of $x$ and coefficients. Errors in cancellation or multiplication can lead to the wrong numerical answer.
• Forgetting the $\cos^3(4x)$ term: The $\cos^3(4x)$ term in the denominator of the first fraction does not go to zero as $x \to 0$, so it should be handled as part of the algebraic simplification, not as a limit that needs manipulation like $\frac{1-\cos(\theta)}{\theta^2}$.

Summary

The problem requires evaluating a limit involving trigonometric functions and logarithms. We first simplify the expression using the identity $1 - \cos^2(\theta) = \sin^2(\theta)$. Then, we strategically rearrange the terms to utilize the standard limits $\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1$ and $\lim_{\theta \to 0} \frac{\log_e(1+\theta)}{\theta} = 1$. By introducing appropriate powers of $x$ in the numerator and denominator, we create these standard forms. The remaining algebraic terms are simplified by canceling out powers of $x$ and evaluating the limit of $\cos^3(4x)$ as $x \to 0$. Multiplying the limits of all the individual parts yields the final answer.

The final answer is $\boxed{18}$.