Problem
(3) $\lim _{x \rightarrow-\infty} \frac{2^{x+1}}{3^{x}}=$ (a) $-\infty$ (b) $\infty$ (c) 1 (d) 0 (e) -1
Solution
Step 1 ：Rewrite the function as a single exponential function: \(\frac{2^{x+1}}{3^x} = \frac{2^x \cdot 2}{3^x}\)
Step 2 ：Find the limit as x approaches negative infinity: \(\lim_{x \to -\infty} \frac{2^x \cdot 2}{3^x} = \boxed{\infty}\)
