If $f(x) = \left[\tan \left(\frac{\pi}{4} + x\right)\right]^{\frac{1}{x}}$ for $x \neq 0$ and $f(x) = k$ for $x = 0$ is continuous at $x = 0$,then $k = \dots$

Examine the following function for continuity: $f(x) = \frac{1}{x-5}, x \neq 5$.

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{x + 2}{x^2 + 3 x + 2}, & x \in R - \{-1, -2\} \\ -1, & x = -2 \\ 0, & x = -1 \end{cases}$ then $f$ is continuous on the set

Let $[\bullet]$ denote the greatest integer function, and let $f(x) = \min \{\sqrt{2}x, x^2\}$. Let $S = \{x \in (-2, 2) : \text{the function } g(x) = |x|[x^2] \text{ is discontinuous at } x\}$. Then $\sum_{x \in S} f(x)$ equals:

$A$ function $f$ from $R$ to $R$ is continuous at a point $a \in R$ if for each $\epsilon > 0$,there exists $\delta > 0$ such that:

If the function $f$ defined on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$ by $f(x)=\begin{cases} \frac{\sqrt{2} \cos x-1}{\cot x-1}, & x \neq \frac{\pi}{4} \\ k, & x=\frac{\pi}{4} \end{cases}$ is continuous,then $k$ is equal to