Official solution
We know that \[\mathop {\lim }\limits_{x \to \infty } \left( {1 + x{1 \over x}} \right) = e\]
\[\therefore\] \[\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)^{2x}} = {e^2}\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left[ {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)\left( {{1 \over {{a \over x} + {b \over {{x^2}}}}}} \right)} \right]^{2x\left( {{a \over x} + {b \over {{x^2}}}} \right)}} = {e^2}\]
\[ \Rightarrow {e^{\mathop {\lim }\limits_{x \to \infty } 2\left[ {a + {b \over x}} \right]}} = {e^2} \Rightarrow {e^{2a}} = e{}^2 \Rightarrow a = 1\]
and \[b \in R\]