If $f: R \to [0, \infty)$ is such that $\lim_{x \to 5} f(x)$ exists and $\lim_{x \to 5} \frac{(f(x))^2 - 9}{\sqrt{|x - 5|}} = 0$,then $\lim_{x \to 5} f(x)$ equals:

The largest value of the non-negative integer $a$ for which $\lim _{x \rightarrow 1}\left\{\frac{-a x+\sin (x-1)+a}{x+\sin (x-1)-1}\right\}^{\frac{1-x}{1-\sqrt{x}}}=\frac{1}{4}$ is

If $\lim _{x \rightarrow 0} \frac{\cos (2 x)+a \cos (4 x)-b}{x^4}$ is finite,then $(a+b)$ is equal to :

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 $\frac{a}{b}$.

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

If $\mathop {\lim }\limits_{x \to 0} \left( {\frac{{3\sin x - 3x + \frac{{{x^3}}}{2}}}{{2{x^n}}}} \right)$ is a finite number,then the greatest value of $n \in N$ is -