If $f(x) = \begin{cases} 4x-5, & x \leq 2 \\ x-k, & x > 2 \end{cases}$ then the value of $k$ for which $\lim_{x \rightarrow 2} f(x)$ exists is equal to:

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 the function $f(x)$ satisfies $\lim_{x \rightarrow 1} \frac{f(x)-2}{x^{2}-1} = \pi$,then $\lim_{x \rightarrow 1} f(x) = $

If $\lim _{x \rightarrow 0}\left\{1+x \log \left(1+a^2\right)\right\}^{1 / x}=2 a \sin ^2 \theta$,where $a>0$ and $\theta \in R$,then:

If $a$ and $b$ are roots of the equation $px^2 + qx + r = 0$,then $\lim_{x \rightarrow b} \frac{1 - \cos 2(px^2 + qx + r)}{2(px - pb)^2}$ is equal to

Let $\beta = \lim_{x \to 0} \frac{\alpha x - (e^{3x} - 1)}{\alpha x(e^{3x} - 1)}$ for some $\alpha \in R$. Then the value of $\alpha + \beta$ is: