Question:
Evaluate the limit:
\[
\lim_{x \to 0} \frac{1 - \cos 4x}{2 \sin^2 x + x \tan 7x}
\]
Evaluate the limit:
\[
\lim_{x \to 0} \frac{1 - \cos 4x}{2 \sin^2 x + x \tan 7x}
\]
Show Hint
When dealing with small angle approximations, expand trigonometric functions using their series expansions and simplify the resulting expressions.
Updated On: Apr 1, 2025
- \( \frac{8}{9} \)
- 0
- \( \frac{9}{8} \)
- \( \infty \)
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The Correct Option is A
Solution and Explanation
We use series expansions of trigonometric functions around 0:
\[
\cos 4x \approx 1 - 8x^2 \quad {and} \quad \sin x \approx x \quad {and} \quad \tan 7x \approx 7x
\]
Substituting these approximations into the expression:
\[
\lim_{x \to 0} \frac{1 - (1 - 8x^2)}{2x^2 + x(7x)} = \frac{8x^2}{2x^2 + 7x^2} = \frac{8}{9}
\]
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