Let $f(x) = \begin{cases} \frac{(1 + \tan x)^{\frac{1}{x}} - e}{x} & x \neq 0 \\ k & x = 0 \end{cases}$ be continuous at $x = 0$,then the value of $k$ is:

For what value of $k$ is the function $f(x) = \begin{cases} \frac{\text{log}(1+2x) \sin x^{\circ}}{x^2}, & x \neq 0 \\ k, & x = 0 \end{cases}$ continuous at $x = 0$?

If $f: R \rightarrow R$ is a function defined by $f(x)=[x-1] \cos \left(\frac{2 x-1}{2}\right) \pi,$ where $[.]$ denotes the greatest integer function,then $f$ is

Let $f(x) = x^2 + x \sin x - \cos x$. Then

Let $f(x) = \begin{cases} \frac{2^x + 2^{3-x} - 6}{\sqrt{2^{-x}} - 2^{1-x}} & \text{if } x > 2 \\ \frac{x^2 - 4}{x - \sqrt{3x - 2}} & \text{if } x < 2 \end{cases}$. Determine the nature of the function at $x = 2$.

If the function given by $f(x) = \left(\frac{4x+1}{1-4x}\right)^{\frac{1}{x}}$ for $x \neq 0$ is continuous at $x = 0$,then the value of $f(0)$ is