If \(f(x)=\begin{cases}a\sin\displaystyle\frac{\pi}{2}(x+1); & x\le 0\\ \displaystyle\frac{\tan x-\sin x}{x^3}; & x\gt 0\end{cases}\) is continuous at \(x=0\), then a is equal to:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{2}\)
For continuity at x = 0, the right-hand limit must equal f(0).
The value at 0 (and the left-hand limit) comes from the first piece:
\( \displaystyle f(0) = a\sin\frac{\pi}{2} = a \)
For the right-hand limit, take \(\tan x\) out of the numerator:
\( \displaystyle \tan x-\sin x = \tan x(1-\cos x) \)
\( \displaystyle \lim_{x\to 0^+}\frac{\tan x-\sin x}{x^3} = \lim_{x\to 0^+}\frac{\tan x}{x}\times\frac{1-\cos x}{x^2} \)
Using \(1-\cos x = 2\sin^2\frac{x}{2}\):
\( \displaystyle \frac{1-\cos x}{x^2} = \frac{1}{2}\left(\frac{\sin(x/2)}{x/2}\right)^2 \to \frac{1}{2} \)
and \(\frac{\tan x}{x} \to 1\), so
\( \displaystyle \lim_{x\to 0^+}\frac{\tan x-\sin x}{x^3} = 1\times\frac{1}{2} = \frac{1}{2} \)
Equating with f(0):
\( \displaystyle a = \frac{1}{2} \)