The value of $\mathop {\lim }\limits_{x \to 0} \frac{1}{x}\left[ {{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{{2x + 1}}} \right) - \frac{\pi }{4}} \right]$ is

Find the limit: $\mathop {\lim }\limits_{x \to 1} (x^3 - x^2 + 1)$

If $\lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha$,then the value of $\frac{\log _e \alpha}{1+\log _e \alpha}$ equals :

If $f(x) = \begin{cases} |x|+1, & x < 0 \\ 0, & x = 0 \\ |x|-1, & x > 0 \end{cases}$,for what value$(s)$ of $a$ does $\lim_{x \to a} f(x)$ exist?

If $a$ is the minimum value of $\sin^2 \theta - \sin \theta + \frac{1}{2}$ and $b = \lim_{x \to \infty} (\sqrt{(x + 1)(x + 2)} - x)$,then $|2a + b| = $

The quadratic polynomial $p(x)$ has roots $1$ and $\alpha$, while quadratic polynomial $q(x)$ has roots $1$ and $\beta$. Let $\alpha$ and $\beta$ be the roots of $r(x) = p(x) + q(x)$. Then $\lim_{x \to \infty} [\sqrt{p(x)} - \sqrt{q(x)}] = $