If $a_1 = 1$ and $a_{n+1} = \frac{4 + 3a_n}{3 + 2a_n}$ for $n \ge 1$,and if $\lim_{n \to \infty} a_n = a$,then the value of $a$ is:

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 $\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1}{x+1}-a x-b\right)=4$,then:

If $\lim _{x \rightarrow 1} \frac{x^2-ax+b}{x-1}=5$,then $(a+b)$ is equal to

Let $\alpha, \beta \in R$ be such that $\lim _{x \rightarrow 0} \frac{x^2 \sin (\beta x)}{\alpha x-\sin x}=1$. Then $6(\alpha+\beta)$ equals

If $\lim _{x \rightarrow 0} \frac{ax-(e^{4x}-1)}{ax(e^{4x}-1)}$ exists and is equal to $b$,then the value of $a-2b$ is ....... .