Tính các giới hạn sau:

a) \[\lim \frac{{{{4.3}^n} + {7^{n + 1}}}}{{{{2.5}^n} + {7^n}}} = \lim \frac{{{{4.3}^n} + {{7.7}^n}}}{{{{2.5}^n} + {7^n}}} = \lim \frac{{4.{{\left( {\frac{3}{7}} \right)}^n} + 7}}{{2.{{\left( {\frac{5}{7}} \right)}^n} + 1}}\]

\[ = \frac{{\lim \left[ {4.{{\left( {\frac{3}{7}} \right)}^n} + 7} \right]}}{{\lim \left[ {2.{{\left( {\frac{5}{7}} \right)}^n} + 1} \right]}} = \frac{{4.\lim {{\left( {\frac{5}{7}} \right)}^n} + \lim 7}}{{2\lim {{\left( {\frac{5}{7}} \right)}^n} + \lim 1}} = \frac{{4.0 + 7}}{{2.0 + 1}} = 7.\]

b) \(\mathop {\lim }\limits_{x \to - {2^ + }} \frac{{\sqrt {{x^2} + 4x + 4} }}{{x + 2}}\)\( = \mathop {\lim }\limits_{x \to - {2^ + }} \frac{{\sqrt {{{\left( {x + 2} \right)}^2}} }}{{x + 2}}\)\( = \mathop {\lim }\limits_{x \to - {2^ + }} \frac{{\left| {x + 2} \right|}}{{x + 2}}\)\( = \mathop {\lim }\limits_{x \to - {2^ + }} \frac{{x + 2}}{{x + 2}} = \mathop {\lim }\limits_{x \to - {2^ + }} 1 = 1.\)

(Vì \(x \to - {2^ + }\) thì \(\left| {x + 2} \right| > 0\) nên \(\left| {x + 2} \right| = x + 2\)).