$\lim _{x \rightarrow \infty}\left(\frac{6 x^2-\cos 3 x}{x^2+5}-\frac{5 x^3+3}{\sqrt{x^6+2}}\right) = $

If $f(x) = \begin{cases} \frac{\sin[x]}{[x]}, & [x] \neq 0 \\ 0, & [x] = 0 \end{cases}$ where $[x]$ denotes the greatest integer less than or equal to $x$,then $\lim_{x \to 0^-} f(x)$ is:

Evaluate the limit: $\lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3$

The limiting value of the function $f(x) = \frac{2\sqrt{2} - (\cos x + \sin x)^3}{1 - \sin 2x}$ as $x \to \frac{\pi}{4}$ is

For $\mathop {\text{Lim}}\limits_{x \to 8} \,\,\frac{{\sin \{ x - 10\} }}{{\{ 10 - x\} }}$ (where $\{ \}$ denotes the fractional part function),determine the existence of the limits.

The value of $\lim _{x \rightarrow 0} \frac{x e^{x}-\sin x}{x}$ is equal to: