The function $f(x) = \frac{x^2 - 1}{x^3 - 1}$ is not defined at $x = 1$. If the function is continuous at $x = 1$,then the value of $f(1)$ will be:

If a real valued function $f(x) = \begin{cases} \frac{x^2+(a+3)x+(a+1)}{x+3} & x \neq -3 \\ -\frac{5}{2} & x = -3 \end{cases}$ is continuous at $x = -3$,then $\lim_{x \rightarrow a} (x^2+x+1) = $

If $f(x) = \begin{cases} \frac{(e^{3x}-1) \sin x^{\circ}}{x^2} & x \neq 0 \\ \frac{\pi}{60} & x = 0 \end{cases}$,then:

If $f(x) = \begin{cases} \frac{\sin 2x}{5x}, & x \ne 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ will be

Define $f: R \to R$ by $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0 \end{cases}$ Then the value of $a$ so that $f$ is continuous at $x = 0$ is:

Find all points of discontinuity of the function $f$ defined by $f(x) = |x| - |x + 1|$.