$$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {1 \over x}} \right)^{2x}}\,\,$$ is equal to

$$\,\,\mathop {Lim}\limits_{x \to \infty } \left( {{{x + \sin x} \over x}} \right)\,\,$$ equal to

With reference to the conventional Cartesian $$(x,y)$$ coordinate system, the vertices of a triangle have the following coordinates: $$\,\left( {{x_1},{y_1}} \right) = \left( {1,0} \right);\,\,\,\left( {{x_2},{y_2}} \right) = \left( {2,2} \right);\,\,\,$$ and $$\,\left( {{x_3},{y_3}} \right) = \left( {4,3} \right).$$ The area of the triangle is equal to

The solution $$\int\limits_0^{\pi /4} {{{\cos }^4}3\theta {{\sin }^3}\,6\theta d\theta \,\,} $$ is :