If $f: R \rightarrow R$ is defined by $f(x) = [x-3] + |x-4|$ for $x \in R$,then $\lim_{x \rightarrow 3^{-}} f(x)$ is equal to

$\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}=$

$\lim _{x \rightarrow 2}\left(\sum_{n=1}^{9} \frac{x}{n(n+1) x^{2}+2(2 n+1) x+4}\right)$ is equal to :

$\lim _{n}$ ${\rightarrow \infty} \frac{\left(1^2-1\right)(n-1)+\left(2^2-2\right)(n-2)+\ldots +\left((n-1)^2-(n-1)\right) \cdot 1}{\left(1^3+2^3+\ldots +n^3\right)-\left(1^2+2^2+\ldots +n^2\right)}$ is equal to:

$\lim _{x \rightarrow \infty}\left(\frac{x+8}{x+1}\right)^{x+5} = \dots$

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