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:

If $\lim _{x \rightarrow 0}\left(\frac{\cos 4 x+a \cos 2 x+b}{x^4}\right)$ is finite,then the values of $a, b$ are respectively :

Define $f(x) = \begin{cases} b - ax & \text{if } x < 2 \\ 3 & \text{if } x = 2 \\ a + 2bx & \text{if } x > 2 \end{cases}$. If $\lim_{x \rightarrow 2} f(x)$ exists,then find the value of $\frac{a}{b}$.

Let $\alpha$ and $\beta$ be the roots of $ax^2 + bx + c = 0$,then $\lim_{x \to \alpha} \frac{1 - \cos(ax^2 + bx + c)}{(x - \alpha)^2}$ is equal to

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

If $\mathop {\lim }\limits_{x \to \infty } \left\{ {\ln \left( {{x^2} + 5x} \right) - 2\ln \left( {cx + 1} \right)} \right\} = -2$,then: