If f(x) = left(frac{1+tan x}{1+sin x}right)^{ | vedclass

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(B) For $f(x)$ to be continuous at $x=0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} \left(\frac{1+	an x}{1+\sin x}ight)^{\operatorna

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(B) For $f(x)$ to be continuous at $x=0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} \left(\frac{1+	an x}{1+\sin x}ight)^{\operatorname{cosec} x} = \lim_{x 	o 0} \exp\left(\operatorname{cosec} x \cdot \ln\left(\frac{1+	an x}{1+\sin x}ight)ight)$.Using the expansion $	an x \approx x + \frac{x^3}{3}$ and $\sin x \approx x - \frac{x^3}{6}$,we have:$\frac{1+	an x}{1+\sin x} \approx \frac{1+x}{1+x} = 1$.Since the limit is of the form $1^\infty$,we use $\lim_{x 	o 0} (1+u)^v = e^{\lim_{x 	o 0} uv}$.$\lim_{x 	o 0} \frac{1}{\sin x} \left(\frac{1+	an x}{1+\sin x} - 1ight) = \lim_{x 	o 0} \frac{1}{\sin x} \left(\frac{	an x - \sin x}{1+\sin x}ight) = \lim_{x 	o 0} \frac{	an x - \sin x}{\sin x(1+\sin x)}$.Using $	an x - \sin x = \sin x(\sec x - 1) \approx x \cdot \frac{x^2}{2} = \frac{x^3}{2}$ and $\sin x(1+\sin x) \approx x(1+x) \approx x$,the limit is $\lim_{x 	o 0} \frac{x^3/2}{x} = 0$.Thus,$f(0) = e^0 = 1$.

If $f(x) = \left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}$ is continuous at $x=0$,then $f(0)$ is equal to

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