If $\mathop {\lim }\limits_{x \to 2} \frac{{\tan \left( {x - 2} \right)\{ {x^2} + (k - 2)x - 2k\} }}{{{x^2} - 4x + 4}} = 5$,then $k$ is equal to

If $\mathop {\lim }\limits_{x \to 1} {\sin ^{ - 1}}\left( {\frac{k}{{\ln x}} - \frac{k}{{x - 1}}} \right)$ exists, then the number of integers in the range of $k$ 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

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:

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 ....... .

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: