What is \lim_{x\rightarrow0-}f(x) equal to?

For the following two (02) items: Let f(x)=\begin{cases}\frac{1-\cos 2x}{x^{2}}&,x \lt 0\\ 0&,x=0\\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}&,x \gt 0\end{cases}

  1. A. 2
  2. B. 4
  3. C. 6
  4. D. 8

Correct Answer: A. 2

Explanation

For x \lt 0, f(x) = \frac{1-\cos 2x}{x^2}. Using the half-angle identity, 1-\cos 2x = 2\sin^2 x. Thus, \lim_{x\to0-} \frac{2\sin^2 x}{x^2} = 2 \lim_{x\to0-} (\frac{\sin x}{x})^2 = 2(1)^2 = 2.

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