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EEJ MAIN Mathematics QUESTION #6948
Question 1

Evaluate: $\displaystyle\lim_{x \to 0} \dfrac{(1-\cos 2x)(3+\cos x)}{x\tan 4x}$

  • $-\dfrac{1}{4}$
  • $\dfrac{1}{2}$
  • 1
  • 2✔️
Correct Answer Explanation

Use standard limits: $\lim_{x\to0}\dfrac{1-\cos2x}{x^2}=2$ and $\lim_{x\to0}\dfrac{\tan4x}{x}=4$.

$= \lim_{x\to0}\dfrac{(1-\cos2x)}{x^2}\cdot\dfrac{(3+\cos x)}{\tan4x/x} = \dfrac{2\cdot(3+1)}{4} = \dfrac{8}{4} = \mathbf{2}$