Answer
$\lim\limits_{x \to 0^+}(1+sin~3x)^{1/x} = e^3$
Work Step by Step
Let $~~y = \lim\limits_{x \to 0^+}(1+sin~3x)^{1/x}$
$ln~y = \lim\limits_{x \to 0^+}ln~(1+sin~3x)^{1/x}$
$ln~y = \lim\limits_{x \to 0^+}\frac{1}{x}~ln~(1+sin~3x)$
$ln~y = \lim\limits_{x \to 0^+}\frac{ln~(1+sin~3x)}{x} = \frac{0}{0}$
We can use L'Hospital's Rule:
$ln~y = \lim\limits_{x \to 0^+}\frac{\frac{3~cos~3x}{1+sin~3x}}{1}$
$ln~y = \lim\limits_{x \to 0^+}\frac{3~cos~3x}{1+sin~3x}$
$ln~y = \frac{3}{1}$
$ln~y = 3$
$y = e^3$
