1. **Problem statement:** We want to find the one-sided limits of the function $$g(x) = \frac{1}{\tan^2(x)}$$ as $x$ approaches 0 from the right ($0^+$) and from the left ($0^-$). 2. **Recall the behavior of $\tan(x)$ near 0:** Since $\tan(x) = \frac{\sin(x)}{\cos(x)}$, and both $\sin(x)$ and $\cos(x)$ are continuous near 0, we know: $$\lim_{x \to 0} \tan(x) = 0$$ 3. **Analyze the sign of $\tan(x)$ near 0:** - For $x \to 0^+$, $\tan(x)$ is positive and close to 0. - For $x \to 0^-$, $\tan(x)$ is negative and close to 0. 4. **Evaluate $g(x) = \frac{1}{\tan^2(x)}$ near 0:** Since $\tan^2(x)$ is the square of $\tan(x)$, it is always non-negative. - As $x \to 0^+$, $\tan^2(x) \to 0^+$, so $$g(x) = \frac{1}{\tan^2(x)} \to +\infty$$ - As $x \to 0^-$, $\tan^2(x) \to 0^+$ as well (because square of negative is positive), so $$g(x) = \frac{1}{\tan^2(x)} \to +\infty$$ 5. **Conclusion:** Both one-sided limits tend to $+\infty$. **Final answer:** $$\lim_{x \to 0^+} g(x) = +\infty \quad \text{and} \quad \lim_{x \to 0^-} g(x) = +\infty$$ This corresponds to option A.