Question 14 Suppose \( \lim _{x \rightarrow 3} f(x)=-3 \) and \( \lim _{x \rightarrow 3} g(x)=-6 \). Find the limit \( \lim _{x \rightarrow 8} \frac{f(x)}{g(x)-2 f(x)} \). Give your answer as an integer or a simplified fraction. If the limit does not exist, enter DNE.

To find the limit \( \lim _{x \rightarrow 8} \frac{f(x)}{g(x) - 2 f(x)} \), we start by recognizing that the limits of the functions \( f(x) \) and \( g(x) \) as \( x \) approaches 3 can help us simplify our expression. Given: \[ \lim_{x \rightarrow 3} f(x) = -3 \] \[ \lim_{x \rightarrow 3} g(x) = -6 \] We first evaluate \( g(x) - 2f(x) \): \[ \lim_{x \rightarrow 3} (g(x) - 2f(x)) = \lim_{x \rightarrow 3} g(x) - 2 \lim_{x \rightarrow 3} f(x) = -6 - 2(-3) = -6 + 6 = 0 \] Now we have: \[ \lim_{x \rightarrow 3} \frac{f(x)}{g(x) - 2f(x)} = \frac{-3}{0} \] This indicates that the function is approaching a form of \(-\frac{3}{0}\), which typically suggests that the limit is undefined or possibly approaches infinity or negative infinity, depending on the direction from which \( g(x) - 2f(x) \) approaches zero. To determine the behavior of \( g(x) - 2f(x) \) around \( x = 3 \), we need to evaluate whether \( g(x) - 2f(x) \) is approaching 0 from the positive or negative side. If we substitute a value slightly less than 3 (let's say 2.9) into \( g(x) \) and \( f(x) \) would result in: - \( g(2.9) \) will be close to \(-6\) (slightly above), - \( f(2.9) \) will be close to \(-3\) leading to \( g(2.9) - 2f(2.9) \) approaching from the positive side (as \( g \text{ plus } 6 \) compensates for twice negative). As \( x \) approaches 3 from the left, \( g(x) - 2f(x) \) approaches 0 positively, which leads: \[ -\frac{3}{0^+} \to -\infty \] Conversely, approaching from the right leads to \( g(x) - 2f(x) \) approaching 0 negatively. Thus, this signifies a split direction. We conclude: \[ \lim_{x \rightarrow 3} \frac{f(x)}{g(x) - 2f(x)} = - \infty \] Thus, the final limit is: \[ \lim_{x \rightarrow 8} \frac{f(x)}{g(x) - 2f(x)} = \text{DNE} \]