Answer
$$\eqalign{
& \left( {\text{a}} \right){e^3} \cr
& \left( {\text{b}} \right)\root 3 \of e \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\mathop {\lim }\limits_{x \to + \infty } {\left( {1 + \frac{3}{x}} \right)^x} \cr
& {\text{Let }}t = \frac{x}{3}{\text{ }} \Rightarrow {\text{ }}x = 3t,{\text{ }}x \to + \infty {\text{ then }}t \to + \infty \cr
& \mathop {\lim }\limits_{x \to + \infty } {\left( {1 + \frac{3}{x}} \right)^x} = \mathop {\lim }\limits_{t \to + \infty } {\left( {1 + \frac{3}{{3t}}} \right)^{3t}} = {\left[ {\mathop {\lim }\limits_{t \to + \infty } {{\left( {1 + \frac{1}{t}} \right)}^t}} \right]^3} \cr
& {\text{From the theorem 6}}{\text{.6}}{\text{.8 we have that }}\mathop {\lim }\limits_{x \to + \infty } {\left( {1 + \frac{1}{x}} \right)^x} = e,{\text{ so}} \cr
& {\left[ {\mathop {\lim }\limits_{t \to + \infty } {{\left( {1 + \frac{1}{t}} \right)}^t}} \right]^3} = {\left( e \right)^3} = {e^3} \cr
& \cr
& \left( {\text{b}} \right)\mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{1/3x}} \cr
& {\text{Rewrite, recall that }}\left( {{a^{mn}}} \right) = {\left( {{a^m}} \right)^n} \cr
& \mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{1/3x}} = {\left[ {\mathop {\lim }\limits_{x \to 0} {{\left( {1 + x} \right)}^{1/x}}} \right]^{1/3}} \cr
& {\text{From the theorem 6}}{\text{.6}}{\text{.8 we have that }}\mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{1/x}} = e,{\text{ so}} \cr
& {\left[ {\mathop {\lim }\limits_{x \to 0} {{\left( {1 + x} \right)}^{1/x}}} \right]^{1/3}} = {\left( e \right)^{1/3}} = \root 3 \of e \cr
& \cr} $$
