Assume that $\lim_{\theta \to 1} f(0)$ exists and $\frac{\theta^2 + 0 - 2}{\theta + 3} \leq \frac{f(0)}{\theta^2} \leq \frac{\theta^2 + 20 - 1}{\theta + 3}$ holds for certain interval containing the point $\theta = -1$ then $\lim_{\theta \to 1} f(0)$ and $\lim_{\theta \to 1} \frac{f(0)}{\theta^2}$ is :
Step-by-Step Solution
Key Concept: Apply the Squeeze Theorem to the inequalities involving f(θ) to find lim_{θ→-1} f(θ)/θ² by evaluating the bounds: (θ²-2)/(θ+3) → (-1)/2 = -1 and (θ²+19)/(θ+3) → 18/2 = -1 as θ→-1, yielding the limit equals -1.
Using the squeeze theorem with the inequalities $-1 \leq \lim_{\theta \to -1} \frac{f(\theta)}{\theta^2} \leq -1$, we conclude that $\lim_{\theta \to -1} \frac{f(\theta)}{\theta^2} = -1$. By algebraic manipulation of the limit expression $\lim_{\theta \to -1} \frac{\theta^2 + 0 - 2}{\theta + 3} = \lim_{\theta \to -1} \frac{\theta^2 - 20 - 1}{\theta + 3}$, we extract the value of the function. Therefore, $\lim_{\theta \to -1} f(\theta) = -1$.
Correct Answer: 1,4