$A$ function is $f(x) = \begin{cases} \frac{e^{1/x}-1}{e^{1/x}+1}, & \text{if } x \neq 0 \\ 0, & \text{if } x=0 \end{cases}$

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{2 \sin x-\sin 2 x}{2 x \cos x}, & \text{if } x \neq 0 \\ a, & \text{if } x=0 \end{cases}$, then the value of $a$ so that $f$ is continuous at $x=0$ is

If $f(x) = \begin{cases} \frac{1-\sqrt{2} \sin x}{\pi-4x} & \text{if } x \neq \frac{\pi}{4} \\ a & \text{if } x = \frac{\pi}{4} \end{cases}$ is continuous at $x = \frac{\pi}{4}$,then $a$ is equal to

If $f(x) = \frac{\log_e(1 + x^2 \tan x)}{\sin x^3}, x \neq 0$ is to be continuous at $x = 0$,then $f(0)$ must be equal to

Let $R$ be the set of all real numbers and $\alpha \in R$ be positive. Define a function $f: R \rightarrow R$ by $f(0)=0$ and $f(x)=|x|^\alpha \sum \limits_{n=0}^{\infty}\left(1+x^2\right)^{-n}$,for $x \neq 0$. Then the set of real numbers $\alpha$ for which $f$ is continuous at $x = 0$ has

For the function $f(x) = \frac{\log_e(1 + x) - \log_e(1 - x)}{x}$ to be continuous at $x = 0$,the value of $f(0)$ should be