$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{{n^2} - n + 1}}{{{n^2} - n - 1}}} \right)^{n(n - 1)}} = $

$\mathop {\lim }\limits_{x \to \infty } \frac{{{{\cot }^{ - 1}}\left( {\sqrt {x + 1} - \sqrt x } \right)}}{{{{\sec }^{ - 1}}\left\{ {{{\left( {\frac{{2x + 1}}{{x - 1}}} \right)}^x}} \right\}}}$ ની કિંમત શોધો.

જો ${x_n} = \frac{{1 - 2 + 3 - 4 + 5 - 6 + \dots - 2n}}{{\sqrt {{n^2} + 1} + \sqrt {4{n^2} - 1} }},$ હોય,તો $\mathop {\lim }\limits_{n \to \infty } {x_n}$ ની કિંમત શોધો.

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \sqrt{2}-(\cos x+\sin x)^3}{1-\sin 2 x}=$

$\lim _{x}$ ${\rightarrow 0} 2\left(\frac{1-\cos x \sqrt{\cos 2 x} \sqrt[3]{\cos 3 x} \ldots \sqrt[10]{\cos 10 x}}{x^2}\right)$ ની કિંમત ............ છે.

જો $\lim _{x \rightarrow 0} \frac{\left(e^{k x}-1\right) \sin k x}{x^{2}}=4$ હોય,તો $k$ ની કિંમત શોધો.