If $\lim _{x \rightarrow 0} \frac{\left(e^{k x}-1\right) \sin k x}{x^{2}}=4$,then $k$ is equal to

$\mathop {\lim }\limits_{x \to \infty } \frac{{2{x^2} - 3x + 1}}{{{x^2} - 1}} = $

If $\lim _{n \rightarrow \infty} x_n$ exists and is finite,$x_1=2$,$x_{n+1}=\frac{a+b x_n}{b+c x_n}$ for all $n \in N$,and $c > b > a > 0$,then $\lim _{n \rightarrow \infty} x_n =$

Evaluate: $\cos \left[ \lim_{x \rightarrow \infty} \frac{2 \pi |x| + \pi x}{|x| - 3x} + \lim_{x \rightarrow 0} \frac{\cos \left( \frac{\pi}{2} \cos^2 x \right)}{x^2} \right]$

$\lim _{x \rightarrow 0}\left[\tan \left(x+\frac{\pi}{4}\right)\right]^{1 / x}$ is equal to

The value of $\lim _{n \rightarrow \infty} n \sin \frac{2 \pi}{3 n} \cos \frac{2 \pi}{3 n}$ is