Standard Limits

If $\text{f}\left(\text{n, }\theta \right)=\underset{\text{r}=1}{\overset{\text{n}}{\Pi }}\left(1-{\text{tan}}^2\frac{\theta }{2^{\text{r}}}\right)$, then compute $\underset{\text{n}\to \infty }{\text{Lim}}\text{ f }\left(\text{n}, \theta \right)$       { given $\theta \ne 0$ }

If $\underset{x\to 0}{\lim }\frac{\sin \left(nx\right)[(a-n)nx-\tan x]}{x^2}=0, (n>0)$ then the value of $\text{'}a\text{'}$ is equal to

If $[.]$ here denotes the greatest integer function, $\underset{x\to 0}{\lim }{x}^7\left[\frac{1}{x^3}\right]=$

$\underset{x\to 0^{+}}{\lim }\frac{x{Sin}^{-1}\left(\frac{2x}{1+x^2}\right)}{{Cos}^{-1}\left(\frac{1-x^2}{1+x^2}\right){Tan}^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)}=$

The limit $\underset{x\to 1}{\lim }\frac{\sqrt{1-\cos 2(x-1)}}{x-1}\_\_\_\_\_\_\_\_\_$

$\underset{x\to (-3)}{\lim }\left(\frac{{\sin }^{-1}(x+3)}{x^2+3x}\right)=$

$\underset{x\to 0}{\lim }\left(\frac{\sin ax}{\tan bx}\right)=$

The value of$\underset{n\to \infty }{\lim }{\sin }^{40}(n!x\pi ),$ where $n\in N$ and $x$ is a rational number, is:

$\underset{x\to 0}{\mathrm{Lim}}\frac{1-{\cos }^{3}x}{x.\sin x.\cos x}$ equals

$\underset{x\to 0}{\lim }\frac{\left(1-\cos 2x\right)\left(3+\cos x\right)}{x\tan 4x}$ is equal to

$\underset{x\to \frac{\pi }{6}}{\lim }\frac{\sin 2x}{\sin x}$ is equal to

$\underset{x\to 0}{\lim }\frac{\tan \left(\left[-{\pi }^2\right]x^2\right)-x^2\tan \left[-{\pi }^2\right]}{{\sin }^{2}x}$ denotes the greatest integers function

$\underset{x\to \frac{\pi }{6}}{\lim }\frac{\sin 2x}{\sin x}$ is equal to

$\underset{x\to -1}{Lim}\frac{\cos 2-\cos 2x}{x^2-\left|x\right|}$ equals

Let $\mathit{\text{f}}\text{: }\mathit{\text{R}}\to \mathit{\text{R}}$ be such that $\mathit{\text{f}}\left(1\right)=2\text{, }{\mathit{\text{f}}}^{\text{'}}\left(1\right)=4$, then find the $\underset{\mathit{\text{x}}\to \mathrm{0 }}{\text{lim}}{\left[\frac{\mathit{\text{f}}\left(1+\mathit{\text{x}}+{\mathit{\text{x}}}^2\right)+\mathit{\text{xf}}\left(1+\mathit{\text{x}}\right)}{\mathit{\text{f}}\left(1\right)}\right]}^{\frac{1}{\text{x}\left(\mathit{\text{x}}+1\right)}}$.

$\underset{x\to 0}{\lim }\frac{a^{x^2}-b^{x^2}}{x^2}=$

Evaluate:$\underset{x\to 0}{\lim }\frac{e^{\sin x}-1}{x}=$

$\underset{x\to 0}{\lim }\frac{e^{2x}-e^{-2x}}{x}=$

$\underset{x\to 0}{\lim }\frac{e^{13x}-e^{7x}}{x}=$