Let \( f(x)=\begin{cases}\displaystyle\frac{1-\cos 4x}{x^2} & , \text{if } x\lt 0\\2\lambda & , \text{if } x=0\\\displaystyle\frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4} & , \text{if } x\gt 0\end{cases} \) such that \( f \) continuous at \( x=0 \) .Then, the value of \( \lambda \) is equal to:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \lambda=4 \)
For continuity at x = 0, the left-hand limit, the right-hand limit and f(0) = 2λ must all be equal.
Left-hand limit, using \( 1-\cos 2\theta=2\sin^2\theta \):
\( \displaystyle \lim_{x\to 0^-}\frac{1-\cos 4x}{x^2} \)
\( \displaystyle =\lim_{x\to 0^-}\frac{2\sin^2 2x}{x^2} \)
\( \displaystyle =\lim_{x\to 0^-}8\left(\frac{\sin 2x}{2x}\right)^2 \)
\( \displaystyle =8 \)
Right-hand limit, multiplying by the conjugate \( \sqrt{16+\sqrt{x}}+4 \):
\( \displaystyle \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4} \)
\( \displaystyle =\frac{\sqrt{x}\left(\sqrt{16+\sqrt{x}}+4\right)}{(16+\sqrt{x})-16} \)
\( \displaystyle =\sqrt{16+\sqrt{x}}+4 \)
\( \displaystyle \lim_{x\to 0^+}\left(\sqrt{16+\sqrt{x}}+4\right)=4+4 \)
\( \displaystyle =8 \)
Both limits are 8, so
\( \displaystyle 2\lambda=8 \)
\( \displaystyle \lambda=4 \)
Tip: A frequent slip is to set \( \lambda \) itself equal to the limit and choose 8; here f(0) is \( 2\lambda \).