1. **State the problem:** We want to find the limit $$\lim_{n \to \infty} \frac{\tan^2 n}{n}.$$\n\n2. **Recall the behavior of the functions:** The numerator is $\tan^2 n$, which oscillates because $\tan n$ is periodic and unbounded near points where $\cos n = 0$. The denominator $n$ grows without bound as $n \to \infty$.\n\n3. **Key insight:** Since $\tan^2 n$ oscillates between $0$ and arbitrarily large values, but the denominator $n$ grows without bound, the fraction $\frac{\tan^2 n}{n}$ will be squeezed by values that become very small as $n$ increases.\n\n4. **Use the squeeze theorem:** For all $n$, $\tan^2 n \geq 0$, so $$0 \leq \frac{\tan^2 n}{n}.$$ Also, since $\tan^2 n$ can be very large but $n$ grows without bound, the fraction tends to zero because the denominator dominates.\n\n5. **Conclusion:** Therefore, $$\lim_{n \to \infty} \frac{\tan^2 n}{n} = 0.$$