Answer
$$\eqalign{
& \left( {\text{a}} \right)\sqrt e \cr
& \left( {\text{b}} \right){e^2} \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\mathop {\lim }\limits_{x \to + \infty } {\left( {1 + \frac{1}{{2x}}} \right)^x} \cr
& {\text{Let }}t = 2x{\text{ }} \Rightarrow {\text{ }}x = \frac{t}{2},{\text{ }}x \to + \infty {\text{ then }}t \to + \infty \cr
& \mathop {\lim }\limits_{x \to + \infty } {\left( {1 + \frac{1}{{2x}}} \right)^x} = \mathop {\lim }\limits_{t \to + \infty } {\left( {1 + \frac{1}{t}} \right)^{t/2}} = {\left[ {\mathop {\lim }\limits_{t \to + \infty } {{\left( {1 + \frac{1}{t}} \right)}^t}} \right]^{1/2}} \cr
& {\text{From the theorem 6}}{\text{.6}}{\text{.8 we have that }}\mathop {\lim }\limits_{x \to + \infty } {\left( {1 + \frac{1}{x}} \right)^x} = e,{\text{ so}} \cr
& {\left[ {\mathop {\lim }\limits_{t \to + \infty } {{\left( {1 + \frac{1}{t}} \right)}^t}} \right]^{1/2}} = {\left( e \right)^{1/2}} = \sqrt e \cr
& \cr
& \left( {\text{b}} \right)\mathop {\lim }\limits_{x \to 0} {\left( {1 + 2x} \right)^{1/x}} \cr
& {\text{Let }}t = 2x{\text{ }} \Rightarrow {\text{ }}x = \frac{t}{2},{\text{ }}x \to 0{\text{ then }}t \to 0 \cr
& \mathop {\lim }\limits_{x \to 0} {\left( {1 + 2x} \right)^{1/x}} = \mathop {\lim }\limits_{t \to 0} {\left( {1 + t} \right)^{2/t}} = {\left[ {\mathop {\lim }\limits_{t \to 0} {{\left( {1 + t} \right)}^{1/t}}} \right]^2} \cr
& {\text{From the theorem 6}}{\text{.6}}{\text{.8 we have that }}\mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{1/x}} = e,{\text{ so}} \cr
& {\left[ {\mathop {\lim }\limits_{t \to 0} {{\left( {1 + t} \right)}^{1/t}}} \right]^2} = {\left( e \right)^2} = {e^2} \cr} $$
