Infinite limit and vertical asymptote at x = -5

As $x \to -5$, the quantity $(x+5)^2 \to 0^+$, since squaring makes it positive regardless of whether $x$ approaches from the left or the right. Since the numerator is the constant $3 > 0$, the fraction $\frac{3}{(x+5)^2}$ becomes a positive number divided by a smaller and smaller positive number.

This means the value of $g(x)$ grows without bound as $x \to -5$ from either side, so $\lim_{x \to -5} g(x) = +\infty$. Since both one-sided limits are infinite (and $g$ is undefined at $x=-5$), the function has a vertical asymptote at $x = -5$.

The function is positive on both sides near x equals negative five. The numerator is the constant three, which is always positive, and the denominator is a squared quantity, which is always positive regardless of direction of approach. So the fraction stays positive on both the left and right sides, which is why the graph shoots upward toward positive infinity from both directions rather than diverging to different signs.