Evaluate: $\mathop {\lim }\limits_{x \to 0} \frac{\sqrt{1+x}-1}{x}$

$\mathop {Lim}\limits_{n \to \infty } \frac{{{1^2}n + {2^2}(n - 1) + {3^2}(n - 2) + \dots + {n^2} \cdot 1}}{{{1^3} + {2^3} + {3^3} + \dots + {n^3}}}$ is equal to :

$\lim _{n \rightarrow \infty}\left(1+\frac{1+\frac{1}{2}+\ldots+\frac{1}{n}}{n^{2}}\right)^{n} = \dots$

$\lim_{x \rightarrow 0} \frac{x \left( e^{\frac{\sqrt{1+x^{2}+x^{4}}-1}{x}} - 1 \right)}{\sqrt{1+x^{2}+x^{4}}-1}$ is equal to:

The value of the limit of $\frac{x^3 - 8}{x^2 - 4}$ as $x \to 2$ is

If $f(x) = \text{Sgn}(\text{Sgn}(\text{Sgn}(x)))$,then $\mathop {\lim }\limits_{x \to 0} f(x)$ is equal to :-