If the function defined by $f(x) = \begin{cases} (x^2 + e^{\frac{1}{2-x}})^{-1}, & x \neq 2 \\ k, & x = 2 \end{cases}$ is right continuous at $x = 2$, then $k =$

Let $f(x) = \begin{cases} x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0 \end{cases}$. Then at $x=0$:

If $f(x) = \begin{cases} x - 1, & x < 0 \\ \frac{1}{4}, & x = 0 \\ x^2, & x > 0 \end{cases}$,then

Given $f(x) = b ([x]^2 + [x]) + 1$ for $x \geq -1$ and $f(x) = \sin(\pi(x+a))$ for $x < -1$,where $[x]$ denotes the greatest integer function,for what values of $a$ and $b$ is the function continuous at $x = -1$?

Let $f(x) = \frac{2 - \sqrt{x + 4}}{\sin 2x}$,$x \neq 0$. In order for $f(x)$ to be continuous at $x = 0$,$f(0)$ must be defined as:

If $f(x) = \frac{e^{x^2} - \cos x}{x^2}$ for $x \neq 0$ is continuous at $x = 0$,then the value of $f(0)$ is