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}
- A. 6
- B. 7
- C. 8 ✓
- D. 9
Correct Answer: C. 8
Explanation
For x \gt 0, rationalise the denominator by multiplying the numerator and denominator by \sqrt{16+\sqrt{x}}+4. The denominator becomes (16+\sqrt{x}) - 16 = \sqrt{x}. The expression simplifies to \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{\sqrt{x}} = \sqrt{16+\sqrt{x}}+4. As x \to 0+, the limit is \sqrt{16+0} + 4 = 4+4=8.
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