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(C) For the function $f(x)$ to be continuous at $x=0$,the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function $f(0)$ must b

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(C) For the function $f(x)$ to be continuous at $x=0$,the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function $f(0)$ must be equal. $f(0) = a$. $LHL$: $\lim_{x 	o 0^-} (1+\sin x)^{\operatorname{cosec} x} = \lim_{x 	o 0^-} (1+\sin x)^{1/\sin x} = e^1 = e$. Since $f(x)$ is continuous at $x=0$,$a = e$. $RHL$: $\lim_{x 	o 0^+} \frac{e^{2/x}+e^{3/x}}{a e^{2/x}+b e^{3/x}}$. Divide numerator and denominator by $e^{3/x}$: $\lim_{x 	o 0^+} \frac{e^{-1/x}+1}{a e^{-1/x}+b} = \frac{0+1}{a(0)+b} = \frac{1}{b}$. Equating $RHL$ to $f(0)$: $\frac{1}{b} = a$. Since $a = e$,we have $\frac{1}{b} = e$,which implies $b = \frac{1}{e} = e^{-1}$. Therefore,$ab = e 	imes e^{-1} = 1$.

If a real valued function $f(x) = \begin{cases} (1+\sin x)^{\operatorname{cosec} x}, & -\pi/2 < x < 0 \\ a, & x=0 \\ \frac{e^{2/x}+e^{3/x}}{a e^{2/x}+b e^{3/x}}, & 0 < x < \pi/2 \end{cases}$ is continuous at $x=0$,then $ab=$

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