$\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1-\sqrt{x}}{1-x}}$ is equal to

If $f(x) = \frac{x(a^x - 1)}{1 - \cos x}$ and $g(x) = \frac{x(1 - a^x)}{a^x(\sqrt{1 - x^2} - \sqrt{1 + x^2})}$,then $\lim_{x \to 0} (f(x) - g(x)) = $

$\lim _{x \rightarrow 0}\left[\tan \left(x+\frac{\pi}{4}\right)\right]^{1 / x}$ is equal to

If $a = \lim_{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$ and $b = \lim_{x \rightarrow 0} \frac{\sin^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$,then the value of $ab^3$ is

$\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x^3-3 x^2+2 x}\right]$ is equal to

The value of $\operatorname{Lim}_{n \rightarrow \infty} \frac{1+2-3+4+5-6+\ldots+(3n-2)+(3n-1)-3n}{\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}}$ is: