MHT CET · Maths · Limits
If \(\quad \mathbf{f}(x)=\left\{\begin{array}{cl}\frac{1-\cos 4 x}{x^2} & \text { if } x < 0 \\ \frac{a}{\sqrt{x}} & \text { if } x=0 \\ \frac{(16+\sqrt{x})^{\frac{1}{2}}-4}{(16+\sqrt{x})} & \text { if } x>0\end{array}\right.\) is continuous at \(x=0\) Then \(a=\)
- A \(4\)
- B \(8\)
- C \(-4\)
- D \(-8\)
Answer & Solution
Correct Answer
(B) \(8\)
Step-by-step Solution
Detailed explanation
\(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1-\cos 4x}{x^2}\) \(\lim_{x \to 0^-} \frac{1-\cos 4x}{x^2} = \frac{4^2}{2}\)
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