Official solution
\[\mathop {\lim }\limits_{x \to \infty } {\left( {{{{x^2} + 5x + 3} \over {{x^2} + x + 2}}} \right)^x}\]
\[ = \mathop {\lim }\limits_{x \to \infty } {\left( {1 + {{4x + 1} \over {{x^2} + x + 2 }}} \right)^x}\]
\[ = \mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 + {{4x + 1} \over {{x^2} + x + 2}}} \right)}^{{{{x^2} + x + 2} \over {4x + 1}}}}} \right]^{{{\left( {4x + 1} \right)x} \over {{x^2} + x + 2}}}}\]
=\[\mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 + {{4x + 1} \over {{x^2} + x + 2}}} \right)}^{{1 \over {{{4x + 1} \over {{x^2} + x + 2}}}}}}} \right]^{{{\left( {4x + 1} \right)x} \over {{x^2} + x + 2}}}}\]
\[ = {e^{\mathop {\lim }\limits_{x \to \infty } {{4{x^2} + x} \over {{x^2} + x + 2}}}}\]
[ As \[\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \lambda x} \right)^{{1 \over x}}} = {e^\lambda }\] ]
\[ = {e^{\mathop {\lim }\limits_{x \to \infty } {{4 + {1 \over x}} \over {1 + {1 \over x} + {2 \over {{x^2}}}}}}} = {e^4}\]