Question

Evaluate the limit: \[ \lim_{x \to 0} \frac{x^2 \sin^2(3x) + \sin^4(6x)}{(1 - \cos(3x))^2} \]

• A. \(\frac{580}{9}\)
• B. \(\frac{145}{3}\)
• C. \(\frac{580}{3}\)
• D. \(\frac{145}{9}\)

Hint:

When evaluating limits of trigonometric functions as \(x \to 0\), it's highly effective to use the standard limits \(\lim_{x \to 0} \frac{\sin(ax)}{ax} = 1\) and \(\lim_{x \to 0} \frac{1 - \cos(ax)}{ax^2} = \frac{a^2}{2}\). Manipulate the expression to isolate these forms. Factor out the highest power of \(x\) from both the numerator and denominator to simplify the expression before applying the limits. This method avoids L'Hôpital's Rule, which can be more complex for such expressions.

Answer 1

The Correct Option is A

Step 1: Use standard limits for trigonometric functions as \(x \to 0\).
We will use the following well-known limits:
1. \(\lim_{x \to 0} \frac{\sin(ax)}{ax} = 1\)
2. \(\lim_{x \to 0} \frac{1 - \cos(ax)}{ax^2} = \frac{a^2}{2}\)
Step 2: Rewrite the numerator using standard limit forms.
The numerator is \(x^2 \sin^2(3x) + \sin^4(6x)\).
For the first term, \(x^2 \sin^2(3x) = x^2 \left(\frac{\sin(3x)}{3x} \cdot 3x\right)^2 = x^2 \left(\frac{\sin(3x)}{3x}\right)^2 (9x^2) = 9x^4 \left(\frac{\sin(3x)}{3x}\right)^2\).
As \(x \to 0\), this term behaves like \(9x^4 (1)^2 = 9x^4\). For the second term, \(\sin^4(6x) = \left(\frac{\sin(6x)}{6x} \cdot 6x\right)^4 = \left(\frac{\sin(6x)}{6x}\right)^4 (6x)^4 = (6x)^4 \left(\frac{\sin(6x)}{6x}\right)^4 = 1296x^4 \left(\frac{\sin(6x)}{6x}\right)^4\).
As \(x \to 0\), this term behaves like \(1296x^4 (1)^4 = 1296x^4\). So, the numerator approximately becomes \(9x^4 + 1296x^4 = 1305x^4\) as \(x \to 0\). Step 3: Rewrite the denominator using standard limit forms. The denominator is \((1 - \cos 3x)^2\).
We know \(\lim_{x \to 0} \frac{1 - \cos(3x)}{(3x)^2} = \frac{1}{2}\). So, \(1 - \cos 3x \approx \frac{1}{2}(3x)^2 = \frac{9x^2}{2}\) as \(x \to 0\).
Therefore, \((1 - \cos 3x)^2 \approx \left(\frac{9x^2}{2}\right)^2 = \frac{81x^4}{4}\) as \(x \to 0\). 
Step 4: Substitute the simplified expressions back into the limit. \[ \lim_{x \to 0} \frac{x^2 \sin^2(3x) + \sin^4(6x)}{(1 - \cos 3x)^2} = \lim_{x \to 0} \frac{9x^4 \left(\frac{\sin(3x)}{3x}\right)^2 + 1296x^4 \left(\frac{\sin(6x)}{6x}\right)^4}{\left(\frac{1 - \cos 3x}{(3x)^2} \cdot (3x)^2\right)^2} \] \[ = \lim_{x \to 0} \frac{x^4 \left[9 \left(\frac{\sin(3x)}{3x}\right)^2 + 1296 \left(\frac{\sin(6x)}{6x}\right)^4\right]}{\left(\frac{1 - \cos 3x}{(3x)^2}\right)^2 (9x^2)^2} \] \[ = \lim_{x \to 0} \frac{x^4 \left[9 \left(\frac{\sin(3x)}{3x}\right)^2 + 1296 \left(\frac{\sin(6x)}{6x}\right)^4\right]}{\left(\frac{1 - \cos 3x}{(3x)^2}\right)^2 (81x^4)} \] Cancel out \(x^4\): \[ = \frac{9 (1)^2 + 1296 (1)^4}{\left(\frac{1}{2}\right)^2 \cdot 81} \] \[ = \frac{9 + 1296}{\frac{1}{4} \cdot 81} \] \[ = \frac{1305}{\frac{81}{4}} \] \[ = 1305 \times \frac{4}{81} \] Now simplify the fraction. Both 1305 and 81 are divisible by 9. \(1305 \div 9 = 145\) \(81 \div 9 = 9\) \[ = \frac{145 \times 4}{9} \] \[ = \frac{580}{9} \]