The function $f(t) = \frac{1}{t^2 + t - 2}$,where $t = \frac{1}{x - 1}$,is discontinuous at

At $x=0$,the function $f(x) = \begin{cases} \frac{x}{|x|+2x^2}, & x \neq 0 \\ k, & x=0 \end{cases}$ is:

If $f(x) = \begin{cases} \frac{x^2 + 3x - 10}{x^2 + 2x - 15}, & x \neq -5 \\ a, & x = -5 \end{cases}$ is continuous at $x = -5$,then the value of $a$ is:

Let $f(x) = \begin{cases} (x - 1)^{\frac{1}{2 - x}}, & x > 1, x \neq 2 \\ k, & x = 2 \end{cases}$. The value of $k$ for which $f$ is continuous at $x = 2$ is

Let $f(x) = \begin{cases} \frac{x - 4}{|x - 4|} + a, & x < 4 \\ a + b, & x = 4 \\ \frac{x - 4}{|x - 4|} + b, & x > 4 \end{cases}$. Then $f(x)$ is continuous at $x = 4$ when

Find the values of $k$ so that the function $f$ is continuous at the indicated point.
$f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & \text{if } x \neq \frac{\pi}{2} \\ 3, & \text{if } x = \frac{\pi}{2} \end{cases}$ at $x = \frac{\pi}{2}$