Question

1. below are the graphs of the functions f and g.

the graph of y = f(x) the graph of y = g(x)
(a) evaluate:
\\(\lim_{x \to 1^-} \frac{-4x^2 - 5x + 2}{f(x)} = \underline{\quad\quad}\\)
\\(\lim_{x \to 1^+} \frac{-4x^2 - 5x + 2}{f(x)} = \underline{\quad\quad}\\)
\\(\lim_{x \to 1} \frac{-4x^2 - 5x + 2}{f(x)} = \underline{\quad\quad}\\)

Explanation:

Step1: Evaluate numerator limit at $x\to1$

First, calculate $\lim_{x\to1} (-4x^2 -5x +2)$:
$$\lim_{x\to1} (-4x^2 -5x +2) = -4(1)^2 -5(1) +2 = -4-5+2 = -7$$
This limit is the same for $x\to1^-$ and $x\to1^+$.

Step2: Find $\lim_{x\to1^-} f(x)$

From the graph of $f(x)$, as $x$ approaches 1 from the left, $f(x)$ approaches 0, and $f(x)<0$ here:
$$\lim_{x\to1^-} f(x) = 0^-$$

Step3: Compute left-hand overall limit

$$\lim_{x\to1^-} \frac{-4x^2 -5x +2}{f(x)} = \frac{-7}{0^-} = +\infty$$

Step4: Find $\lim_{x\to1^+} f(x)$

From the graph of $f(x)$, as $x$ approaches 1 from the right, $f(x)$ approaches 0, and $f(x)>0$ here:
$$\lim_{x\to1^+} f(x) = 0^+$$

Step5: Compute right-hand overall limit

$$\lim_{x\to1^+} \frac{-4x^2 -5x +2}{f(x)} = \frac{-7}{0^+} = -\infty$$

Step6: Check overall limit at $x\to1$

Since $\lim_{x\to1^-} \frac{-4x^2 -5x +2}{f(x)}
eq \lim_{x\to1^+} \frac{-4x^2 -5x +2}{f(x)}$, the two-sided limit does not exist.

Answer:

$\lim_{x\to1^-} \frac{-4x^2 -5x +2}{f(x)} = +\infty$
$\lim_{x\to1^+} \frac{-4x^2 -5x +2}{f(x)} = -\infty$
$\lim_{x\to1} \frac{-4x^2 -5x +2}{f(x)}$ does not exist