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

The value of $\lim_{x \to 0} \frac{\log_{e}(\sec(ex) \cdot \sec(e^{2}x) \cdot ... \cdot \sec(e^{10}x))}{e^{2} - e^{2\cos x}}$ is equal to

$\lim _{x \rightarrow 2} \frac{\sqrt[3]{6+x}-\sqrt[3]{10-x}}{x-2} = $

If $\sum_{r=1}^{n}(2r-1) = x$,then find the value of $\lim_{n}$ ${\rightarrow \infty} \left[ \frac{1^3}{x^2} + \frac{2^3}{x^2} + \frac{3^3}{x^2} + \ldots + \frac{n^3}{x^2} \right]$.

Assertion $(A)$: $\lim _{x \rightarrow 0} \frac{1}{x} = \infty$
Reason $(R)$: As the value of $x$ decreases,the value of $\frac{1}{x}$ increases.

$\mathop {\lim }\limits_{x \to \infty } \,{\left( {\frac{{x + a}}{{x + b}}} \right)^{x + b}} = $