Answer
(a) Please see the figure attached.
The graph suggests that the limit of the sequence may be $3$.
(b) $\mathop {\lim }\limits_{n \to \infty } {\left( {{2^n} + {3^n}} \right)^{1/n}} = 3$
This result confirms our conjecture in part (a).
Work Step by Step
(a) Using a graphing utility we generate the graph of the equation $y = {\left( {{2^x} + {3^x}} \right)^{1/x}}$. Please see the figure attached.
The graph suggests that the limit of the sequence may be $3$.
(b) We evaluate the limit:
$\mathop {\lim }\limits_{n \to \infty } {\left( {{2^n} + {3^n}} \right)^{1/n}} = \mathop {\lim }\limits_{n \to \infty } {\left( {{3^n}\left( {\dfrac{{{2^n}}}{{{3^n}}} + 1} \right)} \right)^{1/n}} = 3\mathop {\lim }\limits_{n \to \infty } {\left( {1 + {{\left( {\dfrac{2}{3}} \right)}^n}} \right)^{1/n}}$
As $n \to \infty $, ${\left( {\dfrac{2}{3}} \right)^n} \to 0$ and $\dfrac{1}{n} \to 0$. Therefore, $\mathop {\lim }\limits_{n \to \infty } {\left( {1 + {{\left( {\dfrac{2}{3}} \right)}^n}} \right)^{1/n}} = 1$. Thus,
$\mathop {\lim }\limits_{n \to \infty } {\left( {{2^n} + {3^n}} \right)^{1/n}} = 3$
This result confirms our conjecture in part (a).
