<s>[INST] $ \lim_{n \to \infty} \frac{(-1)^n(3-n)}{(3n-5)}$ and ... To find limits:
$(a) \lim_{n \to \infty} \frac{(-1)^n(3-n)}{(3n-5)}$
$(b) \lim_{n \to \infty} \frac{n^3}{n!}$

For the first one the sequence is oscillating so it does not converge.
For the 2nd one I used ratio test: Let $a_n = \frac{n^3}{n!} $, then $\frac{a_{n+1}}{a_n} = \frac{n! \times (n+1)^3}{n^3 \times (n+1)!} = (1+ \frac1n)^3 \times \frac{1}{1+n}$. Thus,
$|\frac{a_{n+1}}{a_n}| < 1$, by ratio test limit is $0$.
Is the solutions correct?
 [/INST] Another way for the second is
$$\dfrac{n^3}{n!}<\dfrac{n}{n}\dfrac{n}{n-1}\dfrac{n}{n-2}\dfrac{1}{n-3}<1\dfrac{n}{\frac{n}{2}}\dfrac{n}{\frac{n}{2}}\dfrac{1}{\frac{n}{2}}=\dfrac{8}{n}<\varepsilon$$
for $n>6$.
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