Question

\[ \text{If } \lim_{x \to 0} \frac{\cos 2x - \cos 4x}{1 - \cos 2x} = k, \text{ then evaluate } \lim_{x \to k} \frac{x^k - 27}{x^{k+1} - 81} \]

• A. \( 0 \)
• B. \( 1 \)
• C. \( \dfrac{1}{2} \)
• D. \( \dfrac{1}{4} \)

Hint:

Break complex nested limits into manageable expressions and substitute known values carefully. For rational expressions, rewriting powers as reciprocals helps simplify.

Answer 1

The Correct Option is D

To solve the given problem, we need to evaluate the expression using limits. We start with the first part: \[\lim_{x \to 0} \frac{\cos 2x - \cos 4x}{1 - \cos 2x} = k\]

We use the trigonometric identity for the difference of cosines: \[\cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)\]Apply this identity:

\(-2 \sin(3x) \sin(-x)\ =\ 2 \sin(3x) \sin(x)\)

Now,\[1 - \cos 2x = 2 \sin^2 x\]Substitute these into the limit:

\[\lim_{x \to 0} \frac{2 \sin(3x) \sin x}{2 \sin^2 x} = \lim_{x \to 0} \frac{\sin(3x)}{\sin x} = \lim_{x \to 0} \frac{3 \sin x \cos x}{x\ sin x} = \lim_{x \to 0} 3 \cos x = 3\]

Thus, \(k = 3\).

Next, evaluate: \[\lim_{x \to k} \frac{x^k - 27}{x^{k+1} - 81}\]Substitute \(k = 3\):

\[\lim_{x \to 3} \frac{x^3 - 27}{x^4 - 81}\]

Apply factoring:\[x^3 - 27 = (x - 3)(x^2 + 3x + 9)\]
\[x^4 - 81 = (x - 3)(x^3 + 3x^2 + 9x + 27)\]Cancel \((x - 3)\):

\[\lim_{x \to 3} \frac{x^2 + 3x + 9}{x^3 + 3x^2 + 9x + 27}\]

Substitute \(x = 3\):

\(\frac{3^2 + 3 \cdot 3 + 9}{3^3 + 3 \cdot 3^2 + 9 \cdot 3 + 27} = \frac{9 + 9 + 9}{27 + 27 + 27 + 27} = \frac{27}{108} = \frac{1}{4}\)

The final answer is:

\(\frac{1}{4}\)