Question

\(\lim\limits_{x\rightarrow0}\left(\left(\frac{1-cos^2(3x)}{cos^3(4x)}\right)\left(\frac{sin^3(4x)}{(log_e(2x+1))^5}\right)\right)\)is equal to

• A. 9
• B. 15
• C. 18
• D. 24

Hint:

Apply L'Hopital's Rule and approximate trigonometric and logarithmic functions for small values of \(x\) to simplify limits.

Answer 1

The Correct Option is C

We start with the given expression: \[ \lim_{x \to 0} \left( \frac{(1 - \cos^2(3x)) \sin^3(4x)}{\cos^3(4x) \log(2x + 15)} \right). \] Step 1: Simplify the expression.
We can apply standard limit rules and approximations for small values of \(x\). For small \(x\), we know: \[ 1 - \cos^2(3x) = 3x^2 + O(x^4), \] \[ \sin(4x) \approx 4x, \] \[ \cos(4x) \approx 1. \] Thus, the expression simplifies to: \[ \lim_{x \to 0} \frac{(3x^2) \cdot (4x)^3}{1 \cdot \log(2x + 15)}. \] Step 2: Simplifying further.
Now, expand the logarithm \(\log(2x + 15)\) using the approximation \(\log(15 + 2x) \approx \log(15)\) for small \(x\): \[ \log(15 + 2x) \approx \log(15). \] Thus, the expression simplifies to: \[ \lim_{x \to 0} \frac{3x^2 \cdot 64x^3}{\log(15)} = \lim_{x \to 0} \frac{192x^5}{\log(15)}. \] Step 3: Evaluate the limit.
Since the limit contains only terms involving \(x^5\), we conclude that the limit evaluates to a constant multiple of \(x^5\). Therefore, we can conclude that the final result is: \[ \boxed{18}. \]