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> Question: Find $\lim\_{n \to \infty} \frac{(-1)^n n^{\frac{3}{2}}}{5e^n}$
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My attempt: Since $(-1)^n$ can be written as $\cos(\pi n)$ we can use the squeeze theorem as such: $ -1 \leq \cos(\pi n) \leq 1 \implies \frac{-n^{\frac{3}{2}}}{5e^n} \leq \frac{\cos(\pi n)\cdot n^{\frac{3}{2}}}{5e^n}\leq \frac{n^{\frac{3}{2}}}{5e^n}$ and so we can take the limit as $n$ approaches zero and because exponentials grow faster than polynomials we get: $0\leq \frac{\cos(\pi n)\cdot n^{\frac{3}{2}}}{5e^n}\leq 0$ so $\lim\_{n \to \infty} \frac{(-1)^n n^{\frac{3}{2}}}{5e^n} =0 $.

Is this correct? If so what other ways could I have solved this limit?