The values of the constants $\alpha$ and $\beta$ such that $\lim_{x \to \infty} \left( \frac{x^2 + 1}{x + 1} - \alpha x - \beta \right) = 0$ are respectively:

If $f(x) = \begin{cases} 1+\frac{2x}{a}, & 0 \leq x \leq 1 \\ ax, & 1 < x \leq 2 \end{cases}$,and $\lim_{x \rightarrow 1} f(x)$ exists,then the sum of the cubes of the possible values of $a$ is:

Let $\alpha(a)$ and $\beta(a)$ be the roots of the equation $(\sqrt[3]{1+a}-1) x^2+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$ where $a > -1$. Then $\lim _{a \rightarrow 0^{+}} \alpha(a)$ and $\lim _{a \rightarrow 0^{+}} \beta(a)$ are

Define $f(x) = \begin{cases} \frac{\sqrt{1+px} - \sqrt{1-px}}{x}, & \text{if } -1 \leq x < 0 \\ \frac{2x+1}{x-2}, & \text{if } 0 \leq x \leq 1 \end{cases}$. If $\lim_{x \rightarrow 0} f(x)$ exists,then $p =$

If $\lim _{x \rightarrow 4} \frac{2 x^2+(3+2 a) x+3 a}{x^3-2 x^2-23 x+60}=\frac{11}{9}$,then $\lim _{x \rightarrow a} \frac{x^2+9 x+20}{x^2-x-20}=$

If $\lim_{x \rightarrow 1} \frac{x+x^{2}+x^{3}+\ldots+x^{n}-n}{x-1}=820, (n \in N)$ then the value of $n$ is equal to